Question

The angle of elevation of the top of the tower observed from each of the three points $AB$and $C$ on the ground forming a triangle is the same i.e.$\angle \alpha$. If $R$ is the circumradius of the$∆ABC$, then the height of the tower is

• A. $R\sin \alpha$
• B. $R\cos \alpha$
• C. $Rcot\alpha$
• D. $R\tan \alpha$

Answer 1

The correct option is D

$R\tan \alpha$

Step 1: Given data:

The angle of elevation of the top of the tower was observed from each of the three points$AB$ and $C$ on the ground.

The angle of elevation of three triangles is the same$=\alpha$

$R$ is the circumradius of the$∆ABC$,

Step 2: Draw the required diagram:

From the figure $PQ$is the tower

Step 3: Formula used

The right-angle triangle $\tan \theta =\frac{L\left(Length\right)}{B\left(Base\right)}$

Step 4: Calculate the height of the tower:

Understand from the figure,

$\begin{array}{rcl}\angle PAQ& =& \angle PBQ=\angle PCQ=\alpha \\ \angle PQA& =& \angle PQC=\angle PQB={90}^{\circ }\\ PQ& =& h\end{array}$

Note that in the right-angle triangle APQ,

$\begin{array}{rcl}\tan \alpha & =& \frac{PQ}{AQ}\\ \tan \alpha & =& \frac{h}{AQ}\\ AQ& =& \frac{h}{\tan \alpha }\end{array}$

Now from the right-angle triangles, $BPQ,CPQ$,

$\begin{array}{rcl}BQ& =& \frac{h}{\tan \alpha }\\ CQ& =& \frac{h}{\tan \alpha }\end{array}$

$AQ=BQ=CQ$ (Q is the circumcenter of triangle ABC)

And also $R$is the circumradius of the triangle ABC, $AQ=BQ=CQ=R$

Therefore ,

$$\begin{gathered} R=\frac{h}{\tan \alpha } \\ h=R\tan \alpha \end{gathered}$$

Therefore the height of the tower $R\tan \alpha$.

Hence,option D is the correct answer.