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Mathematics · 2023 · 5 marks
CBSE 2023 · Region 5 · Set 2 · Q34
The angle of elevation of the top of a vertical tower from a point P on the ground is 60^°. From another point Q, 10 m vertically above the first…
The angle of elevation of the top of a vertical tower from a point P on the ground is $\displaystyle 60^{\circ}$. From another point $\displaystyle \mathrm{Q}, 10 \mathrm{~m}$ vertically above the first point P, its angle of elevation is $\displaystyle 30$°. Find :
(a)
The height of the tower.
(b)
The distance of the point P from the foot of the tower.
(c)
The distance of the point P from the top of the tower.
Marking-scheme solution
Let AD be the tower
\[\tan 60^{\circ}=\sqrt{3}=\frac{h}{x} \Rightarrow h=x \sqrt{3}
\]
\(\displaystyle \boldsymbol{\operatorname { t a n }} \mathbf{3 0}^{\boldsymbol{\circ}} \boldsymbol{=} \frac{\mathbf{1}}{\boldsymbol{\sqrt { 3 }}}=\frac{\boldsymbol{h}-\mathbf{1 0}}{\boldsymbol{x}}\)
\[\Rightarrow \mathrm{h}=10+\frac{x}{\sqrt{3}}
\]
\(\displaystyle \boldsymbol{x} \boldsymbol{\sqrt { \mathbf { 3 } }} \boldsymbol{=} \mathbf{1 0} \boldsymbol{+} \frac{\boldsymbol{x}}{\boldsymbol{\sqrt { \mathbf { 3 } }}} \boldsymbol{\Rightarrow} \mathbf{x}=\mathbf{5} \boldsymbol{\sqrt { \mathbf { 3 } }}\)
(a)
\(\displaystyle \mathbf{h} \boldsymbol{=} \mathbf{1 5 ~ m}\)
(b)
\(\displaystyle \mathrm{x}=5 \sqrt{3} \mathrm{~m}\)
(c)
\(\displaystyle \frac{x}{P D}=\cos 60^{\circ} \Rightarrow \mathrm{PD}=10 \sqrt{3} \mathrm{~m}\)
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