Trigonometrical Ratios And Identities Ques 1
- The angle of elevation of the top of a vertical tower standing on a horizontal plane is observed to be $45^{\circ}$ from a point $A$ on the plane. Let $B$ be the point $30 \mathrm{m}$ vertically above the point $A$. If the angle of elevation of the top of the tower from $B$ be $30^{\circ}$, then the distance (in $\mathrm{m}$ ) of the foot of the tower from the point $A$ is
(2019 Main, 12 April II)
(a) $15(3+\sqrt{3})$
(b) $15(5-\sqrt{3})$
(c) $15(3-\sqrt{3})$
(d) $15(1+\sqrt{3})$
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Answer:
Correct Answer: 1.(a)
Solution: (a) According to the question, we have the following figure.
Now, let distance of foot of the tower from the point $A$ is $y \mathrm{m}$.
Draw $B P \perp S T$ such that $P T=x \mathrm{m}$.
Then, in $\triangle T P B$, we have
$ \tan 30^{\circ}=\frac{x}{y} $
$\Rightarrow \quad x=\frac{1}{\sqrt{3}} y$ $\quad$ ……..(i)
and in $\triangle T S A$, we have $\tan 45^{\circ}=\frac{x+30}{y}$
$\Rightarrow \quad y=x+30$ $\quad$ ……..(ii)
On the elimination of quantity $x$ from Eqs. (i) and (ii), we get
$ \begin{aligned} & y=\frac{1}{\sqrt{3}} y+30 \\ & \Rightarrow \quad y\left(1-\frac{1}{\sqrt{3}}\right)=30 \\ & \Rightarrow \quad y=\frac{30 \sqrt{3}}{\sqrt{3}-1}=\frac{30 \sqrt{3}(\sqrt{3}+1)}{3-1} \\ &=\quad \frac{30}{2} \sqrt{3}(\sqrt{3}+1)=15(3+\sqrt{3}) \end{aligned} $
BETA
