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