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Some Applications of Trigonometry
CH09 Question Bank
CBSE_CH09_QUESTION_BANK
Grade 10
Question:
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.
Step-by-Step Solution
Key Concept: 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.
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]
Correct Answer:
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