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Some Applications of Trigonometry
CH09 Question Bank
CBSE_CH09_QUESTION_BANK
Grade 10
Question:
[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]
Step-by-Step Solution
Key Concept: Case study on applications of trigonometry (heights and distances).
(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]
Correct Answer: $\tan\theta=\dfrac{\text{height}}{\text{distance}}$ is the ratio to use, since height is opposite and distance is adjacent to the angle. [1.0 Mark] | $\tan60^\circ=\dfrac{h}{20}\Rightarrow\sqrt3=\dfrac{h}{20}\Rightarrow h=20\sqrt3$ m. [1.0 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] | $\tan\theta=\dfrac{20\sqrt3}{60}=\dfrac{\sqrt3}{3}=\dfrac{1}{\sqrt3}\Rightarrow\theta=30^\circ$, confirming it is indeed smaller than $60^\circ$, consistent with part (c). [1.0 Mark]
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