EXERCISE 9.1

1. Draw the right‑angled triangle formed by the pole, the ground and the rope.\
• Let \(AB\) be the pole (vertical side), \(BC\) the ground (horizontal side) and \(AC\) the rope (hypotenuse).\
• Given: \(AC = 20\,\text{m}\) and \(\angle B = 30^{\circ}\) (angle between rope and ground).\
2. Identify the required side: the height of the pole is \(AB\), which is opposite \(\angle B\).\
3. Use the definition of sine for \(\angle B\):\
$$\sin 30^{\circ} = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{AB}{AC}$$\
4. Substitute the known values:\
$$\sin 30^{\circ} = \frac{AB}{20}\
\Rightarrow AB = 20 \times \sin 30^{\circ}$$\
5. Evaluate \(\sin 30^{\circ} = \frac{1}{2}\):\
$$AB = 20 \times \frac{1}{2} = 10\,\text{m}$$\
6. Hence, the height of the pole is \(10\) metres.

1. Draw a right‑angled triangle:\
- Base (ground) = 8 m (distance from foot of tree to point where top touches ground).\
- Hypotenuse = length of broken part = \(L\).\
- Height = stump height = \(x\).\
- Angle between hypotenuse and ground = \(30^{\circ}\).

2. Apply cosine definition:\
$$\cos 30^{\circ}=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{8}{L}$$\
Hence \(L = \frac{8}{\cos 30^{\circ}}\).

3. Apply sine definition:\
$$\sin 30^{\circ}=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{x}{L}$$\
So \(x = L\sin 30^{\circ}=\frac{L}{2}\).

4. Express original height:\
The original height of the tree \(H\) is the sum of the stump height and the broken part length:\
$$H = x + L = \frac{L}{2}+L = \frac{3L}{2}.$$

5. Compute \(L\):\
\(\cos 30^{\circ}=\frac{\sqrt{3}}{2}\). Therefore\
$$L = \frac{8}{\frac{\sqrt{3}}{2}} = \frac{16}{\sqrt{3}} = \frac{16\sqrt{3}}{3}\;\text{m}.$$

6. Find \(H\):\
$$H = \frac{3}{2}\times \frac{16}{\sqrt{3}} = \frac{24}{\sqrt{3}} = 8\sqrt{3}\;\text{m}.$$

7. Result: The height of the tree is \(8\sqrt{3}\) metres (approximately \(13.9\) m).

1. Draw a right‑angled triangle for each slide:
- The vertical side represents the given height (opposite side).
- The angle between the ground (base) and the slide is the given inclination \(\theta\).
- The slide itself is the hypotenuse of the triangle.

2. Use the sine relation:
$$\sin \theta = \frac{\text{height}}{\text{length of slide}} \quad\Rightarrow\quad \text{length of slide}=\frac{\text{height}}{\sin \theta}$$

3. Case (a): Height = 1.5 m, \(\theta = 30^{\circ}\)
- \(\sin 30^{\circ}=\frac{1}{2}=0.5\)
- $$\text{Length}=\frac{1.5}{0.5}=3\text{ m}$$

4. Case (b): Height = 3 m, \(\theta = 60^{\circ}\)
- \(\sin 60^{\circ}=\frac{\sqrt{3}}{2}\approx 0.866\)
- Exact value:
$$\text{Length}=\frac{3}{\frac{\sqrt{3}}{2}}=\frac{6}{\sqrt{3}}=2\sqrt{3}\text{ m}$$
- Approximate value:
$$\text{Length}\approx \frac{3}{0.866}=3.46\text{ m (to two decimal places)}$$

5. State the answers:
- Slide for children <5 years: 3 m long.
- Slide for elder children: $2\sqrt{3}$ m (≈ 3.46 m) long.

1. Draw a diagram – Let \(AB\) be the tower with \(A\) at the foot and \(B\) at the top. Let \(C\) be the point on the ground 30 m from the foot. Then \(\angle ACB = 30^{\circ}\) (angle of elevation) and \(AC = 30\) m.
2. Identify the right‑angled triangle – \(\triangle ABC\) is a right‑angled triangle with \(\angle BAC = 90^{\circ}\). The side \(AB\) (height of the tower) is opposite the angle \(30^{\circ}\), and \(AC\) is the adjacent side.
3. Apply the tangent definition:
$$\tan 30^{\circ} = \frac{\text{opposite}}{\text{adjacent}} = \frac{AB}{AC}$$
4. Substitute the known values:
$$\tan 30^{\circ} = \frac{AB}{30}\
\Rightarrow AB = 30 \times \tan 30^{\circ}$$
5. Use the exact value \(\tan 30^{\circ} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}\):
$$AB = 30 \times \frac{\sqrt{3}}{3} = 10\sqrt{3}\text{ m}$$
6. If a decimal answer is required, evaluate \(10\sqrt{3} \approx 10 \times 1.732 = 17.32\) m.
7. State the result – The height of the tower is \(10\sqrt{3}\) m (approximately \(17.3\) m).

1. Draw a right‑angled triangle:
- The vertical side (opposite the angle) = height of the kite = $60\,\text{m}$.
- The angle between the string and the ground = $\theta = 60^{\circ}$.
- The string is the hypotenuse (let its length be $\ell$).

2. Apply the sine definition:
$$\sin 60^{\circ} = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{60}{\ell}$$

3. Evaluate $\sin 60^{\circ}$ using the known value $\sin 60^{\circ}=\frac{\sqrt{3}}{2}$:
$$\frac{\sqrt{3}}{2} = \frac{60}{\ell}$$

4. Solve for $\ell$:
$$\ell = \frac{60 \times 2}{\sqrt{3}} = \frac{120}{\sqrt{3}}$$
Rationalise the denominator:
$$\ell = \frac{120}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{120\sqrt{3}}{3}=40\sqrt{3}\,\text{m}$$

5. Numerical value (optional):
$$40\sqrt{3} \approx 40 \times 1.732 = 69.28\,\text{m}$$

Thus, the length of the string is $40\sqrt{3}\,\text{m}$ (approximately $69.3\,\text{m}$).

1. Draw a figure (conceptual):
- Let the foot of the building be point \(B\).
- Let the boy’s eyes be at height \(1.5\,\text{m}\) above the ground.
- Let the top of the building be point \(T\) (height \(30\,\text{m}\)).
- When the boy is at the initial position \(P\), the horizontal distance \(BP = x\) and \(\angle TPB = 30^{\circ}\).
- After walking towards the building to point \(Q\), the new horizontal distance \(BQ = y\) and \(\angle TQB = 60^{\circ}\).

2. Apply the tangent definition for the two positions.
- For position \(P\):
$$\tan 30^{\circ} = \frac{\text{height difference}}{\text{horizontal distance}} = \frac{30-1.5}{x} = \frac{28.5}{x}$$
Hence
$$x = \frac{28.5}{\tan 30^{\circ}}.$$
- For position \(Q\):
$$\tan 60^{\circ} = \frac{28.5}{y}$$
Hence
$$y = \frac{28.5}{\tan 60^{\circ}}.$$

3. Use the known values \(\tan 30^{\circ}=\frac{1}{\sqrt{3}}\) and \(\tan 60^{\circ}=\sqrt{3}\).
- $$x = \frac{28.5}{\frac{1}{\sqrt{3}}}=28.5\sqrt{3}$$
- $$y = \frac{28.5}{\sqrt{3}}=\frac{28.5}{\sqrt{3}}.$$

4. Find the distance walked (the reduction in horizontal distance):
$$\text{Distance walked}=x-y = 28.5\sqrt{3}-\frac{28.5}{\sqrt{3}}
=28.5\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right)
=28.5\left(\frac{3-1}{\sqrt{3}}\right)
=\frac{57}{\sqrt{3}} = \frac{57\sqrt{3}}{3}=19\sqrt{3}\ \text{metres}.$$

5. Numerical value (optional):
$$19\sqrt{3}\approx 19\times1.732 \approx 32.9\ \text{m}.$$

Thus the boy walked approximately 33 metres towards the building.

1. Draw the diagram:\
- Let \(P\) be the point on the ground.\
- Let \(B\) be the foot of the building (ground).\
- Let \(C\) be the top of the building (bottom of the tower) with \(BC = 20\) m.\
- Let \(T\) be the top of the transmission tower, and \(CT = h\) (height of the tower to be found).\
- The horizontal distance \(PB = x\) is common for both observations.
2. Use the 45° angle of elevation to the bottom (point C):\
\[\tan 45^{\circ} = \frac{\text{opposite}}{\text{adjacent}} = \frac{BC}{PB} = \frac{20}{x}.\]
Since \(\tan 45^{\circ}=1\), we get \(x = 20\) m.
3. Use the 60° angle of elevation to the top (point T):\
The total height from ground to T is \(BC + CT = 20 + h\). Hence\
\[\tan 60^{\circ} = \frac{20 + h}{x} = \frac{20 + h}{20}.\]
Because \(\tan 60^{\circ}=\sqrt{3}\),\
\[\sqrt{3} = \frac{20 + h}{20} \quad\Rightarrow\quad 20 + h = 20\sqrt{3}.\]
4. Solve for \(h\):\
\[h = 20\sqrt{3} - 20 = 20(\sqrt{3} - 1)\text{ m}.\]
5. Numerical value (optional):\
\[\sqrt{3} \approx 1.732 \;\Rightarrow\; h \approx 20(1.732 - 1) = 20(0.732) \approx 14.64\text{ m}.\]
Thus the height of the transmission tower is \(20(\sqrt{3}-1)\) m (≈ 14.6 m).

1. Let the height of the pedestal be $h$ metres and the horizontal distance from the observation point to the foot of the pedestal be $x$ metres.
2. From the given angles of elevation:
- For the top of the pedestal, $\tan45^{\circ}=\frac{h}{x}=1 \;\Rightarrow\; h = x$.
- For the top of the statue, whose total height is $h+1.6$, $\tan60^{\circ}=\frac{h+1.6}{x}=\sqrt{3}$.
3. Substitute $x = h$ in the second equation:
$$\frac{h+1.6}{h}=\sqrt{3} \;\Rightarrow\; 1+\frac{1.6}{h}=\sqrt{3}$$
4. Hence $\frac{1.6}{h}=\sqrt{3}-1 \;\Rightarrow\; h=\frac{1.6}{\sqrt{3}-1}$.
5. Rationalising the denominator:
$$h = \frac{1.6(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{1.6(\sqrt{3}+1)}{3-1}=0.8(\sqrt{3}+1)\text{ m}$$
6. Numerically, $h \approx 0.8(1.732+1)=0.8\times2.732 \approx 2.19\text{ m}$.
Thus the height of the pedestal is $0.8(\sqrt{3}+1)\,\text{m}$ (approximately $2.19\,\text{m}$).

1. Let \(d\) be the horizontal distance between the foot of the tower and the foot of the building.\
2. Let \(h\) be the height of the building (required). The tower height is given as \(50\) m.\
3. From the foot of the tower, the angle of elevation to the top of the building is \(30^{\circ}\). Using \(\tan30^{\circ}=\frac{h}{d}\):\
$$\tan30^{\circ}=\frac{h}{d}\quad\Rightarrow\quad \frac{1}{\sqrt{3}}=\frac{h}{d}\quad\Rightarrow\quad h = \frac{d}{\sqrt{3}}.\tag{1}$$\
4. From the foot of the building, the angle of elevation to the top of the tower is \(60^{\circ}\). Using \(\tan60^{\circ}=\frac{50}{d}\):\
$$\tan60^{\circ}=\frac{50}{d}\quad\Rightarrow\quad \sqrt{3}=\frac{50}{d}\quad\Rightarrow\quad d = \frac{50}{\sqrt{3}}.\tag{2}$$\
5. Substitute the value of \(d\) from (2) into equation (1):\
$$h = \frac{\frac{50}{\sqrt{3}}}{\sqrt{3}} = \frac{50}{3}\ \text{metres}.$$
6. Numerically, \(\frac{50}{3} \approx 16.67\) m.

Thus the height of the building is \(\displaystyle \frac{50}{3}\) metres (≈ 16.7 m).

1. Introduce variables
Let the pole whose top subtends an angle of \(60^{\circ}\) be Pole A and the other be Pole B.
Let the distance of the point \(P\) from Pole A be \(x\) metres.
Hence the distance of \(P\) from Pole B is \(80 - x\) metres (since the road width is 80 m).

2. Apply the tangent definition
For Pole A (\(60^{\circ}\)):
$$\tan 60^{\circ}=\frac{h}{x}\quad\Rightarrow\quad \sqrt{3}=\frac{h}{x}\quad\Rightarrow\quad h=\sqrt{3}\,x$$
For Pole B (\(30^{\circ}\)):
$$\tan 30^{\circ}=\frac{h}{80-x}\quad\Rightarrow\quad \frac{1}{\sqrt{3}}=\frac{h}{80-x}\quad\Rightarrow\quad h=\frac{80-x}{\sqrt{3}}$$

3. Equate the two expressions for \(h\)
$$\sqrt{3}\,x = \frac{80-x}{\sqrt{3}}$$
Multiply both sides by \(\sqrt{3}\):
$$3x = 80 - x$$
$$4x = 80$$
$$x = 20\text{ m}$$
Hence the distance from \(P\) to Pole B is \(80 - x = 60\) m.

4. Find the height of the poles
Using \(h = \sqrt{3}\,x\):
$$h = \sqrt{3}\times 20 = 20\sqrt{3}\text{ m} \approx 34.6\text{ m}$$

5. Answer
• Height of each pole = \(20\sqrt{3}\) m (≈ 34.6 m).
• Distance of the point from the pole with \(60^{\circ}\) elevation = 20 m.
• Distance of the point from the pole with \(30^{\circ}\) elevation = 60 m.

1. Introduce variables\
Let
- \(w\) = width of the canal (distance from the point A on the opposite bank to the foot of the tower B).\
- \(h\) = height of the TV tower.

2. First observation (point A)\
From point A the angle of elevation to the top of the tower is \(60^{\circ}\).\
In the right‑angled triangle \(\triangle ABT\) (where T is the top of the tower),\
\[\tan 60^{\circ}=\frac{h}{w}\]
Since \(\tan 60^{\circ}=\sqrt{3}\), we get\
\[\sqrt{3}=\frac{h}{w}\quad\Rightarrow\quad h=\sqrt{3}\,w \tag{1}\]

3. Second observation (point C)\
Point C is 20 m away from A away from the tower along the same straight line (as shown in Fig. 9.12). Hence the distance from C to the foot of the tower is \(w+20\) metres.
The angle of elevation from C is \(30^{\circ}\).\
Using \(\tan 30^{\circ}=\frac{1}{\sqrt{3}}\),\
\[\tan 30^{\circ}=\frac{h}{w+20}\]
gives\
\[\frac{1}{\sqrt{3}}=\frac{h}{w+20}\quad\Rightarrow\quad h=\frac{w+20}{\sqrt{3}} \tag{2}\]

4. Equate the two expressions for \(h\)\
From (1) and (2):\
\[\sqrt{3}\,w = \frac{w+20}{\sqrt{3}}\]
Multiply both sides by \(\sqrt{3}\):\
\[3w = w + 20\]
\[2w = 20\]
\[w = 10\ \text{metres}\]

5. Find the height \(h\)\
Substitute \(w = 10\) into (1):\
\[h = \sqrt{3}\times 10 = 10\sqrt{3}\ \text{metres}\]
Numerically, \(h \approx 17.3\) m.

6. Answer\
- Width of the canal = \(10\) m\
- Height of the TV tower = \(10\sqrt{3}\) m (≈ 17.3 m).

1. Draw a diagram (described):
- Let \(B\) be the top of the building (height = 7 m above ground).
- Let \(F\) be the foot of the tower on the ground.
- Let \(T\) be the top of the tower (height = \(h\) meters above ground).
- The horizontal line through \(B\) meets the ground at a point directly below \(B\); the line \(BF\) makes an angle of depression \(45^{\circ}\) with this horizontal.
- The line of sight \(BT\) makes an angle of elevation \(60^{\circ}\) with the same horizontal.
- Both \(BF\) and \(BT\) share the same horizontal distance \(x = BF = BT_{\text{horizontal}}\).

2. Use the angle of depression (45°) to find the horizontal distance \(x\):
\[
\tan 45^{\circ} = \frac{\text{vertical drop}}{\text{horizontal distance}} = \frac{7}{x}
\]
Since \(\tan 45^{\circ}=1\), we get \(1 = \frac{7}{x}\) ⇒ \(x = 7\) metres.

3. Use the angle of elevation (60°) to relate \(x\) to the height of the tower:
The vertical difference between the tower top and the building top is \(h-7\). Hence,
\[
\tan 60^{\circ} = \frac{h-7}{x}
\]
With \(\tan 60^{\circ}=\sqrt{3}\) and \(x = 7\),
\[
\sqrt{3} = \frac{h-7}{7}
\]
Multiply both sides by 7:
\[
h - 7 = 7\sqrt{3}
\]
Therefore,
\[
h = 7 + 7\sqrt{3} = 7(1+\sqrt{3})\text{ metres}
\]

4. Numerical value (optional):
\(\sqrt{3} \approx 1.732\), so
\[
h \approx 7(1 + 1.732) = 7 \times 2.732 \approx 19.1\text{ m}
\]

Answer: The height of the cable tower is \(7(1+\sqrt{3})\) metres (approximately \(19.1\) m).

1. Draw a diagram: Let $L$ be the top of the lighthouse, $B$ its base on sea level, and $S_1$, $S_2$ the positions of the two ships such that $S_2$ is farther from the lighthouse than $S_1$ and both lie on the same straight line from $B$.
2. Identify right triangles: $\triangle LBS_1$ and $\triangle LBS_2$ are right‑angled at $B$.
3. Use angle of elevation: The angle of depression $\theta$ from $L$ to a ship equals the angle of elevation from the ship to $L$. Hence,
$$\tan\theta = \frac{\text{height of lighthouse}}{\text{horizontal distance from B to the ship}}.$$
4. Compute distances:
- For the ship with angle of depression $45^{\circ}$ (nearest ship):
$$d_1 = \frac{75}{\tan45^{\circ}} = \frac{75}{1} = 75\text{ m}.$$
- For the ship with angle of depression $30^{\circ}$ (farther ship):
$$d_2 = \frac{75}{\tan30^{\circ}} = \frac{75}{\frac{1}{\sqrt3}} = 75\sqrt3\text{ m}.$$
5. Find the distance between the ships:
$$\text{Distance} = d_2 - d_1 = 75\sqrt3 - 75 = 75(\sqrt3 - 1)\text{ m}.$$
6. Numerical value (optional):
$$75(\sqrt3 - 1) \approx 75(1.732 - 1) \approx 75(0.732) \approx 54.9\text{ m}.$$
Thus the two ships are about $55\text{ m}$ apart.

1. Determine the constant vertical distance\
Height of balloon = 88.2 m\
Height of girl = 1.2 m\
Therefore, vertical distance from the girl's eyes to the balloon \(h = 88.2 - 1.2 = 87.0\) m.

2. Let the horizontal distances from the girl to the balloon at the two instants be \(x_1\) (when the angle of elevation is 60°) and \(x_2\) (when the angle becomes 30°).

3. Apply the tangent definition for each position:
- For \(\theta_1 = 60^{\circ}\):
$$\tan 60^{\circ} = \frac{h}{x_1} \quad\Rightarrow\quad \sqrt{3} = \frac{87}{x_1}$$
Hence \(x_1 = \frac{87}{\sqrt{3}} = \frac{87\sqrt{3}}{3} = 29\sqrt{3}\) m.
- For \(\theta_2 = 30^{\circ}\):
$$\tan 30^{\circ} = \frac{h}{x_2} \quad\Rightarrow\quad \frac{1}{\sqrt{3}} = \frac{87}{x_2}$$
Hence \(x_2 = 87\sqrt{3}\) m.

4. Find the horizontal distance travelled by the balloon:
$$\text{Distance} = x_2 - x_1 = 87\sqrt{3} - 29\sqrt{3} = 58\sqrt{3}\ \text{m}.$$

5. Numerical value (optional):
$$58\sqrt{3} \approx 58 \times 1.732 = 100.5\ \text{m}.$$

Thus the balloon travels a horizontal distance of \(58\sqrt{3}\) metres (approximately 100.5 m) during the interval.

1. Let the height of the tower be $h$ metres.
2. At the first observation (angle of depression $30^{\circ}$), the horizontal distance of the car from the foot of the tower is
$$d_1 = \frac{h}{\tan 30^{\circ}} = \frac{h}{\frac{1}{\sqrt3}} = h\sqrt3.$$
3. After 6 s the angle of depression becomes $60^{\circ}$. The new horizontal distance is
$$d_2 = \frac{h}{\tan 60^{\circ}} = \frac{h}{\sqrt3} = \frac{h}{\sqrt3}.$$
4. The car travels the distance $d_1-d_2$ in 6 s, so its uniform speed $v$ is
\begin{align*}
v &= \frac{d_1-d_2}{6}
= \frac{h\sqrt3-\frac{h}{\sqrt3}}{6}
= \frac{h\left(\sqrt3-\frac{1}{\sqrt3}\right)}{6}
= \frac{h\left(\frac{3-1}{\sqrt3}\right)}{6}
= \frac{2h}{6\sqrt3}
= \frac{h}{3\sqrt3}\;\text{m/s}.
\end{align*}
5. The remaining distance to the foot of the tower after the second observation is $d_2 = \frac{h}{\sqrt3}$.
6. Time $t$ required to cover this distance at speed $v$ is
\begin{align*}
t &= \frac{d_2}{v}
= \frac{\frac{h}{\sqrt3}}{\frac{h}{3\sqrt3}}
= \frac{h}{\sqrt3}\times\frac{3\sqrt3}{h}
= 3\;\text{seconds}.
\end{align*}
7. Hence, the car will reach the foot of the tower in 3 seconds from the second observation.

(Alternatively, total time from the first observation is $\frac{d_1}{v}=9$ s; subtracting the elapsed 6 s also gives 3 s.)