Question

The angle of elevation of an object from a point on the level ground is $\alpha$. Moving $d$ meters on the ground towards the object, the angle of elevation is found to be $\beta$. Then, the height (in meters) of the object is

• A. $d\tan \alpha$
• B. $dcot\beta$
• C. $\frac{d}{\left(cot\alpha +cot\beta \right)}$
• D. $\frac{d}{\left(cot\alpha -cot\beta \right)}$

Answer 1

The correct option is D

$\frac{d}{\left(cot\alpha -cot\beta \right)}$

Step 1: Given data:

The angle of elevation of an object from a point on the level ground is $=\alpha$

The angle of elevation after moving $d$ meters on the ground towards the object is found to be $=\beta$

Step 2: Draw the required diagram:

Note that from the figure the height of the object $AB=h$

The angle $\angle ACB=\alpha ,\angle ADB=\beta$

The distance $CD=d,DB=x$

Step 3: Formula used in the right-angle triangle $∆ABD$.

$\tan \theta =\frac{L\left(length\right)}{H\left(Hypotenuse\right)}$

Step 4: Calculate the height of the object:

Note that from the right-angle triangle,

In the $∆ABD$,

$\begin{array}{rcl}\tan \beta & =& \frac{AB}{DB}\\ \tan \beta & =& \frac{h}{x}\\ x& =& hcot\beta ....................\left(1\right)\end{array}$

Now in the right-angle triangle $∆ACB$,

$\begin{array}{rcl}\tan \alpha & =& \frac{AB}{CB}\\ \tan \alpha & =& \frac{h}{x+d}\\ x+d& =& hcot\alpha ....................\left(2\right)\end{array}$

Now substitute the equation (1) from (2).

$\begin{array}{rcl}x+d-x& =& hcot\alpha -hcot\beta \\ d& =& hcot\alpha -hcot\beta \\ h& =& \frac{d}{cot\alpha -cot\beta }\end{array}$

Therefore the height of the object $\frac{d}{\left(cot\alpha -cot\beta \right)}$.

Hence, the correct option is D.