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Some Applications of Trigonometry
CH09 Question Bank
CBSE_CH09_QUESTION_BANK
Grade 10
Question:
[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]
Step-by-Step Solution
Key Concept: Case study on applications of trigonometry (heights and distances).
(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]