CH09 Question Bank
Some Applications of Trigonometry • 50 Questions
Question 1
Hint available
The angle of elevation of an object viewed is the angle formed by the line of sight with the horizontal when the object is:
Above the horizontal level
Below the horizontal level
At the same horizontal level
Behind the observer
Above the horizontal level
Below the horizontal level
At the same horizontal level
Behind the observer
Key Idea
Standard definition of angle of elevation.
Step-by-Step Solution
The angle of elevation is measured upward from the horizontal to the line of sight, when the object is above the observer's eye level. [1.0 Mark]
Question 2
Hint available
The angle of depression of an object viewed is the angle formed by the line of sight with the horizontal when the object is:
Above the horizontal level
Below the horizontal level
At the same horizontal level
Directly overhead
Above the horizontal level
Below the horizontal level
At the same horizontal level
Directly overhead
Key Idea
Standard definition of angle of depression.
Step-by-Step Solution
The angle of depression is measured downward from the horizontal to the line of sight, when the object is below the observer's eye level. [1.0 Mark]
Question 3
Hint available
If an observer looks straight at an object located exactly at their own eye level (the line of sight is horizontal), the angle of elevation is:
$0^\circ$
$45^\circ$
$90^\circ$
Not defined
$0^\circ$
$45^\circ$
$90^\circ$
Not defined
Key Idea
When the line of sight coincides with the horizontal, there is no angle between them.
Step-by-Step Solution
Since the line of sight lies exactly along the horizontal, the angle between them is $0^\circ$. [1.0 Mark]
Question 4
Hint available
In the given figure, a tower $AB$ of height $h$ is observed from point $C$ at a distance $d$, with angle of elevation $\theta$. Which relation correctly connects $h$, $d$, and $\theta$?
$\tan\theta=\dfrac{h}{d}$
$\sin\theta=\dfrac{d}{h}$
$\tan\theta=\dfrac{d}{h}$
$\cos\theta=\dfrac{h}{d}$
$\tan\theta=\dfrac{h}{d}$
$\sin\theta=\dfrac{d}{h}$
$\tan\theta=\dfrac{d}{h}$
$\cos\theta=\dfrac{h}{d}$
Key Idea
The height is opposite to θ and the distance is adjacent to θ; tan relates opposite to adjacent.
Step-by-Step Solution
Since $AB$ (height) is opposite $\theta$ and $BC$ (distance) is adjacent to $\theta$: $\tan\theta=\dfrac{AB}{BC}=\dfrac{h}{d}$. [1.0 Mark]
Question 5
Hint available
A tower stands at a distance of $10\sqrt3$ m from an observer. If the angle of elevation of the top of the tower is $30^\circ$, the height of the tower is:
$5$ m
$10$ m
$10\sqrt3$ m
$15$ m
$5$ m
$10$ m
$10\sqrt3$ m
$15$ m
Key Idea
Apply tanθ = height/distance.
Step-by-Step Solution
$\tan30^\circ=\dfrac{h}{10\sqrt3}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{h}{10\sqrt3}$. [0.5 Mark]
$h=10$ m. [0.5 Mark]
$h=10$ m. [0.5 Mark]
Question 6
Hint available
A pole is observed from a point $20$ m from its base, with an angle of elevation of $45^\circ$. The height of the pole is:
$10$ m
$20$ m
$20\sqrt2$ m
$40$ m
$10$ m
$20$ m
$20\sqrt2$ m
$40$ m
Key Idea
Since tan45° = 1, height equals distance.
Step-by-Step Solution
$\tan45^\circ=\dfrac{h}{20}\Rightarrow 1=\dfrac{h}{20}\Rightarrow h=20$ m. [1.0 Mark]
Question 7
Hint available
A tower is $30$ m high. If the angle of elevation of its top from a point on the ground is $60^\circ$, the distance of the point from the base of the tower is:
$10$ m
$10\sqrt3$ m
$30$ m
$30\sqrt3$ m
$10$ m
$10\sqrt3$ m
$30$ m
$30\sqrt3$ m
Key Idea
Apply tanθ = height/distance and solve for distance.
Step-by-Step Solution
$\tan60^\circ=\dfrac{30}{d}\Rightarrow \sqrt3=\dfrac{30}{d}$. [0.5 Mark]
$d=\dfrac{30}{\sqrt3}=10\sqrt3$ m. [0.5 Mark]
$d=\dfrac{30}{\sqrt3}=10\sqrt3$ m. [0.5 Mark]
Question 8
Hint available
As an observer walks towards the foot of a tower (staying on the same horizontal line), the angle of elevation of the top of the tower:
Increases
Decreases
Remains the same
Becomes zero
Increases
Decreases
Remains the same
Becomes zero
Key Idea
As horizontal distance decreases while height stays fixed, the angle whose tangent is height/distance must increase.
Step-by-Step Solution
Since $\tan\theta=\dfrac{h}{d}$ and $h$ is fixed, decreasing $d$ (moving closer) increases $\tan\theta$, hence $\theta$ increases. [1.0 Mark]
Question 9
Hint available
The angle of elevation of the top of a tower from a point on the ground, and the angle of depression of that same point as seen from the top of the tower, are:
Always equal
Always supplementary
Always complementary
Unrelated
Always equal
Always supplementary
Always complementary
Unrelated
Key Idea
Both angles are formed with parallel horizontal lines and the same line of sight (transversal), making them alternate angles, hence equal.
Step-by-Step Solution
Since the two horizontal lines (at the point and at the top of the tower) are parallel, and the line of sight is a transversal, the angle of elevation and the angle of depression are alternate angles and therefore equal. [1.0 Mark]
Question 10
Hint available
From the top of a tower, the angle of depression of a car on the ground is $30^\circ$. The angle of elevation of the top of the tower as seen from the car is:
$30^\circ$
$60^\circ$
$90^\circ$
Cannot be determined
$30^\circ$
$60^\circ$
$90^\circ$
Cannot be determined
Key Idea
Angle of elevation and angle of depression along the same line of sight are equal (alternate angles).
Step-by-Step Solution
By the alternate-angles property, the angle of elevation from the car equals the angle of depression from the tower, which is $30^\circ$. [1.0 Mark]
Question 11
Hint available
A pole $6$ m tall casts a shadow $6$ m long on the ground. The angle of elevation of the sun at that moment is:
$30^\circ$
$45^\circ$
$60^\circ$
$90^\circ$
$30^\circ$
$45^\circ$
$60^\circ$
$90^\circ$
Key Idea
When height equals shadow length, the tangent of the elevation angle is 1.
Step-by-Step Solution
$\tan\theta=\dfrac{6}{6}=1\Rightarrow\theta=45^\circ$. [1.0 Mark]
Question 12
Hint available
A ladder leaning against a wall makes an angle of $60^\circ$ with the ground. If the ladder is $8$ m long, the height it reaches on the wall is:
$4$ m
$4\sqrt3$ m
$8$ m
$8\sqrt3$ m
$4$ m
$4\sqrt3$ m
$8$ m
$8\sqrt3$ m
Key Idea
The height reached is opposite the angle at the base; the ladder is the hypotenuse.
Step-by-Step Solution
Height $=8\sin60^\circ=8\times\dfrac{\sqrt3}{2}=4\sqrt3$ m. [1.0 Mark]
Question 13
Hint available
The string of a kite makes an angle of $30^\circ$ with the ground. If the kite is flying at a height of $50$ m, the length of the string is:
$50$ m
$50\sqrt3$ m
$100$ m
$100\sqrt3$ m
$50$ m
$50\sqrt3$ m
$100$ m
$100\sqrt3$ m
Key Idea
The height is opposite the angle, and the string is the hypotenuse; use sinθ = height/string length.
Step-by-Step Solution
$\sin30^\circ=\dfrac{50}{\text{string length}}\Rightarrow \dfrac12=\dfrac{50}{L}\Rightarrow L=100$ m. [1.0 Mark]
Question 14
Hint available
As per the CBSE syllabus, problems on heights and distances should not involve more than:
One right triangle
Two right triangles
Three right triangles
Any number of right triangles
One right triangle
Two right triangles
Three right triangles
Any number of right triangles
Key Idea
Direct recall of the official scope restriction for this topic.
Step-by-Step Solution
The syllabus restricts such problems to at most two right triangles, with angles of elevation/depression limited to $30^\circ, 45^\circ, 60^\circ$. [1.0 Mark]
Question 15
Hint available
From the top of a cliff $60$ m high, the angle of depression of a boat on the sea is $60^\circ$. The distance of the boat from the base of the cliff is:
$20$ m
$20\sqrt3$ m
$60$ m
$60\sqrt3$ m
$20$ m
$20\sqrt3$ m
$60$ m
$60\sqrt3$ m
Key Idea
Apply tanθ = height/distance, using the fact that the angle of depression equals the angle of elevation from the boat.
Step-by-Step Solution
$\tan60^\circ=\dfrac{60}{d}\Rightarrow \sqrt3=\dfrac{60}{d}$. [0.5 Mark]
$d=\dfrac{60}{\sqrt3}=20\sqrt3$ m. [0.5 Mark]
$d=\dfrac{60}{\sqrt3}=20\sqrt3$ m. [0.5 Mark]
Question 16
Hint available
Assertion (A): If the angle of elevation of the sun changes from $30^\circ$ to $60^\circ$, the length of a pole's shadow decreases.
Reason (R): For a fixed pole height, the shadow length is $h\cot\theta$, and $\cot\theta$ decreases as $\theta$ increases.
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Reason (R): For a fixed pole height, the shadow length is $h\cot\theta$, and $\cot\theta$ decreases as $\theta$ increases.
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Key Idea
As the sun rises higher (larger elevation angle), shadows shrink.
Step-by-Step Solution
As the elevation angle increases, shadows become shorter (a well-known everyday observation, e.g. shadows are longest at sunrise/sunset and shortest at midday) — A is true. [0.5 Mark]
R correctly gives the formula (shadow $=h\cot\theta$) and correctly notes $\cot\theta$ decreases as $\theta$ increases, which directly explains why the shadow shortens — R correctly explains A. [0.5 Mark]
R correctly gives the formula (shadow $=h\cot\theta$) and correctly notes $\cot\theta$ decreases as $\theta$ increases, which directly explains why the shadow shortens — R correctly explains A. [0.5 Mark]
Question 17
Hint available
Assertion (A): A tower's angle of elevation from a point $20$ m away is $45^\circ$, so the height of the tower is $20$ m.
Reason (R): $\tan 45^\circ = 1$.
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Reason (R): $\tan 45^\circ = 1$.
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Key Idea
Direct computation using the standard value of tan 45°.
Step-by-Step Solution
$\tan45^\circ=\dfrac{h}{20}$. Since $\tan45^\circ=1$, $h=20$ m — A is true. [0.5 Mark]
R correctly states the exact standard value used in this computation, so R correctly explains A. [0.5 Mark]
R correctly states the exact standard value used in this computation, so R correctly explains A. [0.5 Mark]
Question 18
Hint available
Assertion (A): The angle of elevation of the top of a tower from a point on the ground is always the same as the angle of depression of that same point from the top of the tower.
Reason (R): Both angles are measured from the vertical line joining the top and bottom of the tower.
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Reason (R): Both angles are measured from the vertical line joining the top and bottom of the tower.
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Key Idea
A is a correct standard fact, but R gives the wrong justification — the correct reasoning uses horizontal lines and alternate angles, not the vertical line.
Step-by-Step Solution
This equality is a standard, correct property (proved using alternate angles), so A is true. [0.5 Mark]
R is incorrect: the equality arises because both angles are measured from parallel HORIZONTAL lines, with the line of sight as a transversal (giving equal alternate angles) — not because they are measured from the vertical line. So R is false. [0.5 Mark]
R is incorrect: the equality arises because both angles are measured from parallel HORIZONTAL lines, with the line of sight as a transversal (giving equal alternate angles) — not because they are measured from the vertical line. So R is false. [0.5 Mark]
Question 19
Hint available
Assertion (A): A ladder of length $10$ m makes an angle of $30^\circ$ with the ground; it reaches a height of $5$ m on the wall.
Reason (R): Height reached $=\text{ladder length}\times\sin(\text{angle with ground})$.
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Reason (R): Height reached $=\text{ladder length}\times\sin(\text{angle with ground})$.
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Key Idea
The height on the wall is opposite the ground angle, and the ladder is the hypotenuse.
Step-by-Step Solution
Height $=10\sin30^\circ=10\times\dfrac12=5$ m — A is true. [0.5 Mark]
R correctly states the formula used (height $=$ ladder length $\times\sin$ of the angle with the ground), which is exactly the method used to verify A — R correctly explains A. [0.5 Mark]
R correctly states the formula used (height $=$ ladder length $\times\sin$ of the angle with the ground), which is exactly the method used to verify A — R correctly explains A. [0.5 Mark]
Question 20
Hint available
Assertion (A): Heights and distances problems in the current CBSE syllabus may use angles of elevation or depression of any measure, such as $20^\circ$ or $50^\circ$.
Reason (R): The official syllabus restricts such problems to angles of $30^\circ$, $45^\circ$, and $60^\circ$ only.
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Reason (R): The official syllabus restricts such problems to angles of $30^\circ$, $45^\circ$, and $60^\circ$ only.
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Key Idea
Syllabus-awareness check on the specific angle restriction for this chapter.
Step-by-Step Solution
The syllabus explicitly restricts problems to angles of $30^\circ$, $45^\circ$, and $60^\circ$ — arbitrary angles like $20^\circ$ or $50^\circ$ are not used, so A is false. [0.5 Mark]
R correctly states this exact restriction, so R is true. [0.5 Mark]
R correctly states this exact restriction, so R is true. [0.5 Mark]
Question 21
Hint available
The angle of elevation of the top of a tower from a point $15$ m away from its base is $60^\circ$. Find the height of the tower.
Key Idea
Apply tanθ = height/distance.
Step-by-Step Solution
$\tan60^\circ=\dfrac{h}{15}\Rightarrow \sqrt3=\dfrac{h}{15}$. [1.0 Mark]
$h=15\sqrt3$ m. [1.0 Mark]
$h=15\sqrt3$ m. [1.0 Mark]
Question 22
Hint available
A tree is $12$ m tall. Find the distance of a point on the ground from the base of the tree, at which the angle of elevation of the top of the tree is $45^\circ$.
Key Idea
Apply tanθ = height/distance.
Step-by-Step Solution
$\tan45^\circ=\dfrac{12}{d}\Rightarrow 1=\dfrac{12}{d}$. [1.0 Mark]
$d=12$ m. [1.0 Mark]
$d=12$ m. [1.0 Mark]
Question 23
Hint available
A pole casts a shadow of length equal to its own height. Find the angle of elevation of the sun at that time.
Key Idea
When height equals shadow length, tanθ = 1.
Step-by-Step Solution
Let height $=$ shadow $=x$. Then $\tan\theta=\dfrac{x}{x}=1$. [1.0 Mark]
$\theta=45^\circ$. [1.0 Mark]
$\theta=45^\circ$. [1.0 Mark]
Question 24
Hint available
From the top of a building $50$ m high, the angle of depression of a car on the ground is $30^\circ$. Find the distance of the car from the base of the building.
Key Idea
The angle of depression equals the angle of elevation from the car (alternate angles); apply tanθ = height/distance.
Step-by-Step Solution
$\tan30^\circ=\dfrac{50}{d}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{50}{d}$. [1.0 Mark]
$d=50\sqrt3$ m. [1.0 Mark]
$d=50\sqrt3$ m. [1.0 Mark]
Question 25
Hint available
A ladder $6$ m long is placed against a wall, making an angle of $45^\circ$ with the ground. Find the height it reaches on the wall.
Key Idea
The height reached is opposite the ground angle; the ladder is the hypotenuse.
Step-by-Step Solution
Height $=6\sin45^\circ=6\times\dfrac{1}{\sqrt2}$. [1.0 Mark]
$=\dfrac{6}{\sqrt2}=3\sqrt2$ m. [1.0 Mark]
$=\dfrac{6}{\sqrt2}=3\sqrt2$ m. [1.0 Mark]
Question 26
Hint available
A kite is flying at a height of $30\sqrt3$ m, attached to a string making an angle of $60^\circ$ with the ground. Find the length of the string.
Key Idea
Apply sinθ = height/string length.
Step-by-Step Solution
$\sin60^\circ=\dfrac{30\sqrt3}{L}\Rightarrow \dfrac{\sqrt3}{2}=\dfrac{30\sqrt3}{L}$. [1.0 Mark]
$L=\dfrac{30\sqrt3\times2}{\sqrt3}=60$ m. [1.0 Mark]
$L=\dfrac{30\sqrt3\times2}{\sqrt3}=60$ m. [1.0 Mark]
Question 27
Hint available
An observer $1.5$ m tall is $28.5$ m away from a tower $30$ m high (measuring from the observer's eye level). Find the angle of elevation of the top of the tower from the observer's eye.
Key Idea
First find the effective height of the tower above eye level, then apply tanθ = effective height/distance.
Step-by-Step Solution
Effective height above eye level $=30-1.5=28.5$ m. [1.0 Mark]
$\tan\theta=\dfrac{28.5}{28.5}=1\Rightarrow\theta=45^\circ$. [1.0 Mark]
$\tan\theta=\dfrac{28.5}{28.5}=1\Rightarrow\theta=45^\circ$. [1.0 Mark]
Question 28
Hint available
The angle of elevation of the top of a tower from a point on the ground is $30^\circ$. If the height of the tower is $50$ m, find the distance of the point from the base of the tower.
Key Idea
Apply tanθ = height/distance.
Step-by-Step Solution
$\tan30^\circ=\dfrac{50}{d}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{50}{d}$. [1.0 Mark]
$d=50\sqrt3$ m. [1.0 Mark]
$d=50\sqrt3$ m. [1.0 Mark]
Question 29
Hint available
A vertical stick $20$ m long casts a shadow $20\sqrt3$ m long on the ground. Find the angle of elevation of the sun.
Key Idea
Apply tanθ = height/shadow length.
Step-by-Step Solution
$\tan\theta=\dfrac{20}{20\sqrt3}=\dfrac{1}{\sqrt3}$. [1.0 Mark]
$\theta=30^\circ$. [1.0 Mark]
$\theta=30^\circ$. [1.0 Mark]
Question 30
Hint available
From the top of a cliff $100$ m high, the angle of depression of a boat is $45^\circ$. Find the distance of the boat from the base of the cliff.
Key Idea
Use the equal alternate angle property, then apply tanθ = height/distance.
Step-by-Step Solution
$\tan45^\circ=\dfrac{100}{d}\Rightarrow 1=\dfrac{100}{d}$. [1.0 Mark]
$d=100$ m. [1.0 Mark]
$d=100$ m. [1.0 Mark]
Question 31
Hint available
Explain, with reference to a right triangle, why the angle of elevation of an object increases as an observer walks towards its base.
Key Idea
Use the tanθ = height/distance relation to explain the trend.
Step-by-Step Solution
For a fixed height $h$, $\tan\theta=\dfrac{h}{d}$, where $d$ is the horizontal distance from the base. [1.0 Mark]
As the observer walks closer, $d$ decreases, so $\dfrac{h}{d}$ increases, meaning $\tan\theta$ (and hence $\theta$) increases. [1.0 Mark]
As the observer walks closer, $d$ decreases, so $\dfrac{h}{d}$ increases, meaning $\tan\theta$ (and hence $\theta$) increases. [1.0 Mark]
Question 32
Hint available
A $15$ m long ladder makes an angle of $60^\circ$ with the wall (NOT the ground). Find the height it reaches on the wall.
Key Idea
Since the angle is measured from the wall (vertical), the height reached is adjacent to this angle.
Step-by-Step Solution
Height $=15\cos60^\circ$ (adjacent side, using the angle from the vertical wall). [1.0 Mark]
$=15\times\dfrac12=7.5$ m. [1.0 Mark]
$=15\times\dfrac12=7.5$ m. [1.0 Mark]
Question 33
Hint available
The angle of elevation of the top of a tower from a point $40$ m away from its base is $30^\circ$. Find the height of the tower, giving your answer both in surd form and correct to one decimal place (using $\sqrt3\approx1.73$).
Key Idea
Apply tanθ = height/distance, then rationalise and approximate.
Step-by-Step Solution
$\tan30^\circ=\dfrac{h}{40}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{h}{40}$. [1.0 Mark]
$h=\dfrac{40}{\sqrt3}=\dfrac{40\sqrt3}{3}$ m (in surd form). [1.0 Mark]
Using $\sqrt3\approx1.73$: $h\approx\dfrac{40\times1.73}{3}=\dfrac{69.2}{3}\approx23.1$ m. [1.0 Mark]
$h=\dfrac{40}{\sqrt3}=\dfrac{40\sqrt3}{3}$ m (in surd form). [1.0 Mark]
Using $\sqrt3\approx1.73$: $h\approx\dfrac{40\times1.73}{3}=\dfrac{69.2}{3}\approx23.1$ m. [1.0 Mark]
Question 34
Hint available
A $1.6$ m tall observer is $20$ m away from a building. The angle of elevation of the top of the building from the observer's eyes is $60^\circ$. Find the height of the building.
Key Idea
Find the height of the building above eye level first, then add the observer's own height.
Step-by-Step Solution
Let the height above eye level be $x$: $\tan60^\circ=\dfrac{x}{20}\Rightarrow\sqrt3=\dfrac{x}{20}$. [1.0 Mark]
$x=20\sqrt3$ m. [1.0 Mark]
Total height of building $=x+1.6=(20\sqrt3+1.6)$ m $\approx35.2+1.6=36.8$ m (using $\sqrt3\approx1.76$ approx, or leave in surd form as $(20\sqrt3+1.6)$ m). [1.0 Mark]
$x=20\sqrt3$ m. [1.0 Mark]
Total height of building $=x+1.6=(20\sqrt3+1.6)$ m $\approx35.2+1.6=36.8$ m (using $\sqrt3\approx1.76$ approx, or leave in surd form as $(20\sqrt3+1.6)$ m). [1.0 Mark]
Question 35
Hint available
From a point on the ground, the angle of elevation of the top of a $10\sqrt3$ m tall tower is observed. If the point is $10$ m from the base of the tower, find the angle of elevation.
Key Idea
Apply tanθ = height/distance and identify the standard angle.
Step-by-Step Solution
$\tan\theta=\dfrac{10\sqrt3}{10}=\sqrt3$. [1.5 Marks]
$\theta=60^\circ$. [1.5 Marks]
$\theta=60^\circ$. [1.5 Marks]
Question 36
Hint available
Two poles of equal height stand on either side of a road $80$ m wide. From a point on the road between the poles, the angles of elevation of the top of the poles are $30^\circ$ and $60^\circ$. Find the height of the poles and the distance of the point from each pole. (Assume the point is directly between the two poles, on the line joining their bases.)
Key Idea
Let the point be at distance x from one pole and (80-x) from the other; set up two tan equations using the same height, and solve.
Step-by-Step Solution
Let the height of each pole be $h$, and let the point be at distance $x$ from the pole with elevation $60^\circ$, so it is at distance $(80-x)$ from the pole with elevation $30^\circ$. [0.5 Mark]
$\tan60^\circ=\dfrac{h}{x}\Rightarrow h=x\sqrt3$ — (1). $\tan30^\circ=\dfrac{h}{80-x}\Rightarrow h=\dfrac{80-x}{\sqrt3}$ — (2). [1.0 Mark]
Equating (1) and (2): $x\sqrt3=\dfrac{80-x}{\sqrt3}\Rightarrow 3x=80-x\Rightarrow 4x=80\Rightarrow x=20$ m. [1.0 Mark]
$h=20\sqrt3$ m; the point is $20$ m from one pole and $60$ m from the other. [0.5 Mark]
$\tan60^\circ=\dfrac{h}{x}\Rightarrow h=x\sqrt3$ — (1). $\tan30^\circ=\dfrac{h}{80-x}\Rightarrow h=\dfrac{80-x}{\sqrt3}$ — (2). [1.0 Mark]
Equating (1) and (2): $x\sqrt3=\dfrac{80-x}{\sqrt3}\Rightarrow 3x=80-x\Rightarrow 4x=80\Rightarrow x=20$ m. [1.0 Mark]
$h=20\sqrt3$ m; the point is $20$ m from one pole and $60$ m from the other. [0.5 Mark]
Question 37
Hint available
A vertical pole is $15$ m high, and its shadow is $5\sqrt3$ m long. Find the angle of elevation of the sun at that time.
Key Idea
Apply tanθ = height/shadow.
Step-by-Step Solution
$\tan\theta=\dfrac{15}{5\sqrt3}$. [1.0 Mark]
$=\dfrac{15}{5\sqrt3}=\dfrac{3}{\sqrt3}=\sqrt3$. [1.0 Mark]
$\theta=60^\circ$. [1.0 Mark]
$=\dfrac{15}{5\sqrt3}=\dfrac{3}{\sqrt3}=\sqrt3$. [1.0 Mark]
$\theta=60^\circ$. [1.0 Mark]
Question 38
Hint available
From the top of a $75$ m high lighthouse, the angle of depression of a boat is $45^\circ$. Some time later, the boat has moved closer, and the new angle of depression is $60^\circ$. Find the distance the boat travelled.
Key Idea
Find the two horizontal distances separately using the two elevation angles (equal to the depression angles), then subtract.
Step-by-Step Solution
At $45^\circ$: $\tan45^\circ=\dfrac{75}{d_1}\Rightarrow d_1=75$ m. [1.0 Mark]
At $60^\circ$: $\tan60^\circ=\dfrac{75}{d_2}\Rightarrow d_2=\dfrac{75}{\sqrt3}=25\sqrt3$ m. [1.0 Mark]
Distance travelled $=d_1-d_2=75-25\sqrt3=25(3-\sqrt3)$ m $\approx25(3-1.73)=25(1.27)\approx31.75$ m. [1.0 Mark]
At $60^\circ$: $\tan60^\circ=\dfrac{75}{d_2}\Rightarrow d_2=\dfrac{75}{\sqrt3}=25\sqrt3$ m. [1.0 Mark]
Distance travelled $=d_1-d_2=75-25\sqrt3=25(3-\sqrt3)$ m $\approx25(3-1.73)=25(1.27)\approx31.75$ m. [1.0 Mark]
Question 39
Hint available
A ladder rests against a vertical wall such that its foot is $6$ m from the wall and it makes an angle of $60^\circ$ with the ground. Find the length of the ladder and the height it reaches on the wall.
Key Idea
Use the given angle with cos to find the ladder length, and sin to find the height.
Step-by-Step Solution
$\cos60^\circ=\dfrac{6}{L}\Rightarrow \dfrac12=\dfrac{6}{L}\Rightarrow L=12$ m. [1.5 Marks]
Height $=L\sin60^\circ=12\times\dfrac{\sqrt3}{2}=6\sqrt3$ m. [1.5 Marks]
Height $=L\sin60^\circ=12\times\dfrac{\sqrt3}{2}=6\sqrt3$ m. [1.5 Marks]
Question 40
Hint available
A tower stands vertically on the ground. From a point on the ground $20$ m away from the foot of the tower, the angle of elevation of the top is found to be $60^\circ$. Find the height of the tower correct to two decimal places (use $\sqrt3=1.732$).
Key Idea
Apply tanθ = height/distance, then substitute the given decimal approximation.
Step-by-Step Solution
$\tan60^\circ=\dfrac{h}{20}\Rightarrow\sqrt3=\dfrac{h}{20}$. [1.0 Mark]
$h=20\sqrt3$. [1.0 Mark]
$h=20\times1.732=34.64$ m. [1.0 Mark]
$h=20\sqrt3$. [1.0 Mark]
$h=20\times1.732=34.64$ m. [1.0 Mark]
Question 41
Hint available
The shadow of a tower standing on level ground is found to be $40$ m longer when the sun's altitude (angle of elevation) is $30^\circ$ than when it is $60^\circ$. Find the height of the tower.
Key Idea
Set up two equations for the two shadow lengths using the same height, subtract to eliminate the unknown shadow length at 60°.
Step-by-Step Solution
Let the height be $h$ and the shorter shadow (at $60^\circ$) be $s$. At $60^\circ$: $\tan60^\circ=\dfrac{h}{s}\Rightarrow s=\dfrac{h}{\sqrt3}$. [1.0 Mark]
At $30^\circ$, the shadow is $s+40$: $\tan30^\circ=\dfrac{h}{s+40}\Rightarrow s+40=h\sqrt3$. [1.0 Mark]
Substituting: $\dfrac{h}{\sqrt3}+40=h\sqrt3\Rightarrow 40=h\sqrt3-\dfrac{h}{\sqrt3}=h\left(\sqrt3-\dfrac{1}{\sqrt3}\right)=h\times\dfrac{2}{\sqrt3}\Rightarrow h=20\sqrt3$ m. [1.0 Mark]
At $30^\circ$, the shadow is $s+40$: $\tan30^\circ=\dfrac{h}{s+40}\Rightarrow s+40=h\sqrt3$. [1.0 Mark]
Substituting: $\dfrac{h}{\sqrt3}+40=h\sqrt3\Rightarrow 40=h\sqrt3-\dfrac{h}{\sqrt3}=h\left(\sqrt3-\dfrac{1}{\sqrt3}\right)=h\times\dfrac{2}{\sqrt3}\Rightarrow h=20\sqrt3$ m. [1.0 Mark]
Question 42
Hint available
A man standing on the deck of a ship, $8$ m above sea level, observes the angle of elevation of the top of a cliff as $45^\circ$ and the angle of depression of the base of the cliff as $30^\circ$. Find the distance of the cliff from the ship.
Key Idea
Use the angle of depression (equal to elevation from the base) to find the horizontal distance directly.
Step-by-Step Solution
The man's eye level is $8$ m above sea level, and the base of the cliff is at sea level, so the angle of depression to the base gives: $\tan30^\circ=\dfrac{8}{d}$, where $d$ is the horizontal distance. [1.5 Marks]
$\dfrac{1}{\sqrt3}=\dfrac{8}{d}\Rightarrow d=8\sqrt3$ m. [1.5 Marks]
$\dfrac{1}{\sqrt3}=\dfrac{8}{d}\Rightarrow d=8\sqrt3$ m. [1.5 Marks]
Question 43
Hint available
A flagstaff stands on top of a $20$ m high building. From a point on the ground, the angle of elevation of the bottom of the flagstaff (top of the building) is $30^\circ$, and the angle of elevation of the top of the flagstaff is $60^\circ$. Find the height of the flagstaff.
Key Idea
This uses exactly two right triangles sharing the same horizontal distance: one to the top of the building, one to the top of the flagstaff. Find the distance from the first triangle, then use it in the second.
Step-by-Step Solution
Let the observer be at distance $d$ from the base of the building, and let the flagstaff's height be $f$. [0.5 Mark]
Using the angle to the top of the building ($30^\circ$): $\tan30^\circ=\dfrac{20}{d}\Rightarrow\dfrac{1}{\sqrt3}=\dfrac{20}{d}\Rightarrow d=20\sqrt3$ m. [1.5 Marks]
Using the angle to the top of the flagstaff ($60^\circ$), the total height is $20+f$: $\tan60^\circ=\dfrac{20+f}{d}\Rightarrow\sqrt3=\dfrac{20+f}{20\sqrt3}$. [1.5 Marks]
$20+f=20\sqrt3\times\sqrt3=60\Rightarrow f=40$ m. [1.5 Marks]
Using the angle to the top of the building ($30^\circ$): $\tan30^\circ=\dfrac{20}{d}\Rightarrow\dfrac{1}{\sqrt3}=\dfrac{20}{d}\Rightarrow d=20\sqrt3$ m. [1.5 Marks]
Using the angle to the top of the flagstaff ($60^\circ$), the total height is $20+f$: $\tan60^\circ=\dfrac{20+f}{d}\Rightarrow\sqrt3=\dfrac{20+f}{20\sqrt3}$. [1.5 Marks]
$20+f=20\sqrt3\times\sqrt3=60\Rightarrow f=40$ m. [1.5 Marks]
Question 44
Hint available
A tower stands vertically on the ground. From a point on the ground, which is $15$ m away from the foot of the tower, the angle of elevation of the top of the tower is found to be $60^\circ$. From another point $D$, further along the same straight line, the angle of elevation is $30^\circ$. Find the height of the tower and the distance $CD$ between the two observation points $C$ and $D$.
Key Idea
Two right triangles share the same tower height; use the nearer point to find the height, then the farther point to find the additional distance.
Step-by-Step Solution
Let the height of the tower be $h$. From point $C$ (distance $15$ m): $\tan60^\circ=\dfrac{h}{15}\Rightarrow\sqrt3=\dfrac{h}{15}$. [1.0 Mark]
$h=15\sqrt3$ m. [1.0 Mark]
From point $D$ (distance $15+CD$): $\tan30^\circ=\dfrac{h}{15+CD}\Rightarrow\dfrac{1}{\sqrt3}=\dfrac{15\sqrt3}{15+CD}$. [1.5 Marks]
$15+CD=15\sqrt3\times\sqrt3=45\Rightarrow CD=30$ m. [1.5 Marks]
$h=15\sqrt3$ m. [1.0 Mark]
From point $D$ (distance $15+CD$): $\tan30^\circ=\dfrac{h}{15+CD}\Rightarrow\dfrac{1}{\sqrt3}=\dfrac{15\sqrt3}{15+CD}$. [1.5 Marks]
$15+CD=15\sqrt3\times\sqrt3=45\Rightarrow CD=30$ m. [1.5 Marks]
Question 45
Hint available
From the top of a $50$ m high building, the angle of elevation of the top of a tower is $60^\circ$, and the angle of depression of the foot of the tower is $30^\circ$. Find the height of the tower and the distance between the building and the tower.
Key Idea
Split into two right triangles: one using the depression angle to find the horizontal distance (the two structures' bases lie on the same horizontal level), and another using the elevation angle to find the extra height of the tower above the building's height.
Step-by-Step Solution
Let $d$ be the horizontal distance between the building and the tower, and let the tower's height be $H$. Using the angle of depression to the tower's foot ($30^\circ$), and noting this angle equals the angle of elevation of the top of the building as seen from the tower's foot: $\tan30^\circ=\dfrac{50}{d}\Rightarrow\dfrac{1}{\sqrt3}=\dfrac{50}{d}$. [1.5 Marks]
$d=50\sqrt3$ m. [1.0 Mark]
Using the angle of elevation to the top of the tower ($60^\circ$) from the top of the building, the extra height above the building's own height is $x$: $\tan60^\circ=\dfrac{x}{d}\Rightarrow\sqrt3=\dfrac{x}{50\sqrt3}$. [1.0 Mark]
$x=50\sqrt3\times\sqrt3=150$ m. Total height of tower $H=50+150=200$ m. [1.5 Marks]
$d=50\sqrt3$ m. [1.0 Mark]
Using the angle of elevation to the top of the tower ($60^\circ$) from the top of the building, the extra height above the building's own height is $x$: $\tan60^\circ=\dfrac{x}{d}\Rightarrow\sqrt3=\dfrac{x}{50\sqrt3}$. [1.0 Mark]
$x=50\sqrt3\times\sqrt3=150$ m. Total height of tower $H=50+150=200$ m. [1.5 Marks]
Question 46
Hint available
Two poles of heights $6$ m and $11$ m stand vertically on a level ground. The angle of elevation of the top of the taller pole, as observed from the top of the shorter pole, is $30^\circ$. Find the distance between the two poles.
Key Idea
Draw a horizontal line from the top of the shorter pole to the taller pole; the vertical rise (height difference) and the horizontal distance between the poles form a single right triangle with the given angle.
Step-by-Step Solution
Height difference between the poles $=11-6=5$ m. [1.5 Marks]
Drawing a horizontal line from the top of the shorter pole, a right triangle is formed with the vertical leg $=5$ m (height difference) and the horizontal leg $=$ the distance between the poles, $d$. [1.5 Marks]
$\tan30^\circ=\dfrac{5}{d}\Rightarrow\dfrac{1}{\sqrt3}=\dfrac{5}{d}$. [1.0 Mark]
$d=5\sqrt3$ m $\approx8.66$ m. [1.0 Mark]
Drawing a horizontal line from the top of the shorter pole, a right triangle is formed with the vertical leg $=5$ m (height difference) and the horizontal leg $=$ the distance between the poles, $d$. [1.5 Marks]
$\tan30^\circ=\dfrac{5}{d}\Rightarrow\dfrac{1}{\sqrt3}=\dfrac{5}{d}$. [1.0 Mark]
$d=5\sqrt3$ m $\approx8.66$ m. [1.0 Mark]
Question 47
Hint available
[Case Study]
A hot air balloon operator wants to estimate the height of a tall tree before launch. She stands at a point $20$ m from the base of the tree and measures the angle of elevation of the top of the tree to be $60^\circ$.
(a) Which trigonometric ratio directly relates the height of the tree, the distance, and the angle of elevation here? [1 Mark]
(b) Find the height of the tree. [1 Mark]
(c) If the operator moves to a point $60$ m from the base instead, will the angle of elevation be larger or smaller than $60^\circ$? [1 Mark]
(d) Find the new angle of elevation at $60$ m distance, using the height found in part (b). [1 Mark]
A hot air balloon operator wants to estimate the height of a tall tree before launch. She stands at a point $20$ m from the base of the tree and measures the angle of elevation of the top of the tree to be $60^\circ$.
(a) Which trigonometric ratio directly relates the height of the tree, the distance, and the angle of elevation here? [1 Mark]
(b) Find the height of the tree. [1 Mark]
(c) If the operator moves to a point $60$ m from the base instead, will the angle of elevation be larger or smaller than $60^\circ$? [1 Mark]
(d) Find the new angle of elevation at $60$ m distance, using the height found in part (b). [1 Mark]
Key Idea
Case study on applications of trigonometry (heights and distances).
Step-by-Step Solution
(a) Which trigonometric ratio directly relates the height of the tree, the distance, and the angle of elevation here? [1 Mark]
$\tan\theta=\dfrac{\text{height}}{\text{distance}}$ is the ratio to use, since height is opposite and distance is adjacent to the angle. [1.0 Mark]
(b) Find the height of the tree. [1 Mark]
$\tan60^\circ=\dfrac{h}{20}\Rightarrow\sqrt3=\dfrac{h}{20}\Rightarrow h=20\sqrt3$ m. [1.0 Mark]
(c) If the operator moves to a point $60$ m from the base instead, will the angle of elevation be larger or smaller than $60^\circ$? [1 Mark]
Since the distance has increased while the height stays the same, $\tan\theta=h/d$ decreases, so the angle of elevation will be smaller than $60^\circ$. [1.0 Mark]
(d) Find the new angle of elevation at $60$ m distance, using the height found in part (b). [1 Mark]
$\tan\theta=\dfrac{20\sqrt3}{60}=\dfrac{\sqrt3}{3}=\dfrac{1}{\sqrt3}\Rightarrow\theta=30^\circ$, confirming it is indeed smaller than $60^\circ$, consistent with part (c). [1.0 Mark]
$\tan\theta=\dfrac{\text{height}}{\text{distance}}$ is the ratio to use, since height is opposite and distance is adjacent to the angle. [1.0 Mark]
(b) Find the height of the tree. [1 Mark]
$\tan60^\circ=\dfrac{h}{20}\Rightarrow\sqrt3=\dfrac{h}{20}\Rightarrow h=20\sqrt3$ m. [1.0 Mark]
(c) If the operator moves to a point $60$ m from the base instead, will the angle of elevation be larger or smaller than $60^\circ$? [1 Mark]
Since the distance has increased while the height stays the same, $\tan\theta=h/d$ decreases, so the angle of elevation will be smaller than $60^\circ$. [1.0 Mark]
(d) Find the new angle of elevation at $60$ m distance, using the height found in part (b). [1 Mark]
$\tan\theta=\dfrac{20\sqrt3}{60}=\dfrac{\sqrt3}{3}=\dfrac{1}{\sqrt3}\Rightarrow\theta=30^\circ$, confirming it is indeed smaller than $60^\circ$, consistent with part (c). [1.0 Mark]
Question 48
Hint available
[Case Study]
A lighthouse keeper standing at the top of a lighthouse $60$ m high observes a fishing boat at sea. The angle of depression of the boat from the top of the lighthouse is $45^\circ$. As the boat moves closer to the shore, the angle of depression changes to $60^\circ$.
(a) Find the initial distance of the boat from the foot of the lighthouse when the angle of depression was $45^\circ$. [1 Mark]
(b) Find the new distance of the boat from the foot of the lighthouse when the angle of depression became $60^\circ$. [1 Mark]
(c) Calculate the distance travelled by the boat towards the lighthouse during this observation. [1 Mark]
(d) If the boat continues to move towards the lighthouse, will the angle of depression increase or decrease? [1 Mark]
A lighthouse keeper standing at the top of a lighthouse $60$ m high observes a fishing boat at sea. The angle of depression of the boat from the top of the lighthouse is $45^\circ$. As the boat moves closer to the shore, the angle of depression changes to $60^\circ$.
(a) Find the initial distance of the boat from the foot of the lighthouse when the angle of depression was $45^\circ$. [1 Mark]
(b) Find the new distance of the boat from the foot of the lighthouse when the angle of depression became $60^\circ$. [1 Mark]
(c) Calculate the distance travelled by the boat towards the lighthouse during this observation. [1 Mark]
(d) If the boat continues to move towards the lighthouse, will the angle of depression increase or decrease? [1 Mark]
Key Idea
Case study on applications of trigonometry (heights and distances).
Step-by-Step Solution
(a) Find the initial distance of the boat from the foot of the lighthouse when the angle of depression was $45^\circ$. [1 Mark]
$\tan45^\circ = \dfrac{60}{d_1} \Rightarrow 1 = \dfrac{60}{d_1} \Rightarrow d_1 = 60$ m. [1.0 Mark]
(b) Find the new distance of the boat from the foot of the lighthouse when the angle of depression became $60^\circ$. [1 Mark]
$\tan60^\circ = \dfrac{60}{d_2} \Rightarrow \sqrt3 = \dfrac{60}{d_2} \Rightarrow d_2 = \dfrac{60}{\sqrt3} = 20\sqrt3$ m. [1.0 Mark]
(c) Calculate the distance travelled by the boat towards the lighthouse during this observation. [1 Mark]
$d_1 - d_2 = 60 - 20\sqrt3 = 20(3 - \sqrt3)$ m. [1.0 Mark]
(d) If the boat continues to move towards the lighthouse, will the angle of depression increase or decrease? [1 Mark]
As the boat moves closer to the base, the distance $d$ decreases, so $\tan\theta = h/d$ increases, hence the angle of depression increases. [1.0 Mark]
$\tan45^\circ = \dfrac{60}{d_1} \Rightarrow 1 = \dfrac{60}{d_1} \Rightarrow d_1 = 60$ m. [1.0 Mark]
(b) Find the new distance of the boat from the foot of the lighthouse when the angle of depression became $60^\circ$. [1 Mark]
$\tan60^\circ = \dfrac{60}{d_2} \Rightarrow \sqrt3 = \dfrac{60}{d_2} \Rightarrow d_2 = \dfrac{60}{\sqrt3} = 20\sqrt3$ m. [1.0 Mark]
(c) Calculate the distance travelled by the boat towards the lighthouse during this observation. [1 Mark]
$d_1 - d_2 = 60 - 20\sqrt3 = 20(3 - \sqrt3)$ m. [1.0 Mark]
(d) If the boat continues to move towards the lighthouse, will the angle of depression increase or decrease? [1 Mark]
As the boat moves closer to the base, the distance $d$ decreases, so $\tan\theta = h/d$ increases, hence the angle of depression increases. [1.0 Mark]
Question 49
Hint available
[Case Study]
A surveyor is measuring the height of a transmission tower standing vertically on top of a hill. From a point on level ground, the angle of elevation of the bottom of the tower (top of the hill) is $30^\circ$, and the angle of elevation of the top of the tower is $60^\circ$. The distance from the point of observation to the base of the hill is $90$ m.
(a) Find the height of the hill. [1 Mark]
(b) Find the total height from the ground to the top of the transmission tower. [1 Mark]
(c) Find the height of the transmission tower alone. [1 Mark]
(d) If $\sqrt3 \approx 1.732$, calculate the height of the transmission tower in metres correct to one decimal place. [1 Mark]
A surveyor is measuring the height of a transmission tower standing vertically on top of a hill. From a point on level ground, the angle of elevation of the bottom of the tower (top of the hill) is $30^\circ$, and the angle of elevation of the top of the tower is $60^\circ$. The distance from the point of observation to the base of the hill is $90$ m.
(a) Find the height of the hill. [1 Mark]
(b) Find the total height from the ground to the top of the transmission tower. [1 Mark]
(c) Find the height of the transmission tower alone. [1 Mark]
(d) If $\sqrt3 \approx 1.732$, calculate the height of the transmission tower in metres correct to one decimal place. [1 Mark]
Key Idea
Case study on applications of trigonometry (heights and distances).
Step-by-Step Solution
(a) Find the height of the hill. [1 Mark]
$\tan30^\circ = \dfrac{h_{\text{hill}}}{90} \Rightarrow \dfrac{1}{\sqrt3} = \dfrac{h_{\text{hill}}}{90} \Rightarrow h_{\text{hill}} = \dfrac{90}{\sqrt3} = 30\sqrt3$ m. [1.0 Mark]
(b) Find the total height from the ground to the top of the transmission tower. [1 Mark]
$\tan60^\circ = \dfrac{H_{\text{total}}}{90} \Rightarrow \sqrt3 = \dfrac{H_{\text{total}}}{90} \Rightarrow H_{\text{total}} = 90\sqrt3$ m. [1.0 Mark]
(c) Find the height of the transmission tower alone. [1 Mark]
$h_{\text{tower}} = 90\sqrt3 - 30\sqrt3 = 60\sqrt3$ m. [1.0 Mark]
(d) If $\sqrt3 \approx 1.732$, calculate the height of the transmission tower in metres correct to one decimal place. [1 Mark]
$60 \times 1.732 = 103.92 \approx 103.9$ m. [1.0 Mark]
$\tan30^\circ = \dfrac{h_{\text{hill}}}{90} \Rightarrow \dfrac{1}{\sqrt3} = \dfrac{h_{\text{hill}}}{90} \Rightarrow h_{\text{hill}} = \dfrac{90}{\sqrt3} = 30\sqrt3$ m. [1.0 Mark]
(b) Find the total height from the ground to the top of the transmission tower. [1 Mark]
$\tan60^\circ = \dfrac{H_{\text{total}}}{90} \Rightarrow \sqrt3 = \dfrac{H_{\text{total}}}{90} \Rightarrow H_{\text{total}} = 90\sqrt3$ m. [1.0 Mark]
(c) Find the height of the transmission tower alone. [1 Mark]
$h_{\text{tower}} = 90\sqrt3 - 30\sqrt3 = 60\sqrt3$ m. [1.0 Mark]
(d) If $\sqrt3 \approx 1.732$, calculate the height of the transmission tower in metres correct to one decimal place. [1 Mark]
$60 \times 1.732 = 103.92 \approx 103.9$ m. [1.0 Mark]
Question 50
Hint available
[Case Study]
A kite enthusiast is flying a kite at a park. The kite string is stretched taut and makes an angle of $60^\circ$ with the horizontal ground. The length of the string released is $100$ m. Simultaneously, a drone hovering at a fixed height observes the kite.
(a) Find the vertical height of the kite above the ground. [1 Mark]
(b) Find the horizontal distance of the kite from the person holding the string. [1 Mark]
(c) Verify the Pythagorean relationship $h^2 + d^2 = L^2$ for this kite position. [1 Mark]
(d) If the string angle decreases to $30^\circ$ while maintaining the string length at $100$ m, what will be the new height of the kite? [1 Mark]
A kite enthusiast is flying a kite at a park. The kite string is stretched taut and makes an angle of $60^\circ$ with the horizontal ground. The length of the string released is $100$ m. Simultaneously, a drone hovering at a fixed height observes the kite.
(a) Find the vertical height of the kite above the ground. [1 Mark]
(b) Find the horizontal distance of the kite from the person holding the string. [1 Mark]
(c) Verify the Pythagorean relationship $h^2 + d^2 = L^2$ for this kite position. [1 Mark]
(d) If the string angle decreases to $30^\circ$ while maintaining the string length at $100$ m, what will be the new height of the kite? [1 Mark]
Key Idea
Case study on applications of trigonometry (heights and distances).
Step-by-Step Solution
(a) Find the vertical height of the kite above the ground. [1 Mark]
$\sin60^\circ = \dfrac{h}{100} \Rightarrow \dfrac{\sqrt3}{2} = \dfrac{h}{100} \Rightarrow h = 50\sqrt3$ m. [1.0 Mark]
(b) Find the horizontal distance of the kite from the person holding the string. [1 Mark]
$\cos60^\circ = \dfrac{d}{100} \Rightarrow \dfrac{1}{2} = \dfrac{d}{100} \Rightarrow d = 50$ m. [1.0 Mark]
(c) Verify the Pythagorean relationship $h^2 + d^2 = L^2$ for this kite position. [1 Mark]
$h^2 + d^2 = (50\sqrt3)^2 + 50^2 = 7500 + 2500 = 10000 = 100^2 = L^2$. Verified. [1.0 Mark]
(d) If the string angle decreases to $30^\circ$ while maintaining the string length at $100$ m, what will be the new height of the kite? [1 Mark]
$h_{\text{new}} = 100 \sin30^\circ = 100 \times \dfrac{1}{2} = 50$ m. [1.0 Mark]
$\sin60^\circ = \dfrac{h}{100} \Rightarrow \dfrac{\sqrt3}{2} = \dfrac{h}{100} \Rightarrow h = 50\sqrt3$ m. [1.0 Mark]
(b) Find the horizontal distance of the kite from the person holding the string. [1 Mark]
$\cos60^\circ = \dfrac{d}{100} \Rightarrow \dfrac{1}{2} = \dfrac{d}{100} \Rightarrow d = 50$ m. [1.0 Mark]
(c) Verify the Pythagorean relationship $h^2 + d^2 = L^2$ for this kite position. [1 Mark]
$h^2 + d^2 = (50\sqrt3)^2 + 50^2 = 7500 + 2500 = 10000 = 100^2 = L^2$. Verified. [1.0 Mark]
(d) If the string angle decreases to $30^\circ$ while maintaining the string length at $100$ m, what will be the new height of the kite? [1 Mark]
$h_{\text{new}} = 100 \sin30^\circ = 100 \times \dfrac{1}{2} = 50$ m. [1.0 Mark]