Question

From the top of a cliff 200 meters high, the angles of depression of the top and bottom of a tower are
observed to be $30{}^{\circ }$ and $60{}^{\circ }$ . Find the height of the tower and calculate the distance
between them.
A. Height = 156 m; Distance = 119.7 m
B. Height = $133{\displaystyle \frac{1}{3}}$ m; Distance = 115.46 m
C. Height = 220 m; Distance = 112.76 m
D. None of these.

Answer 1

Hint:If a person stands and looks up at an object, the angle of elevation is the angle between the horizontal line of sight and the object. If a person stands and looks down at an object, the angle of depression is the angle between the horizontal line of sight and the object.

Draw a diagram with the positions of the given structures with respect to the level ground.
Draw some right-angled triangles and identify the given angles in it.
Recall the values of trigonometric ratios for $30{}^{\circ }$ and $60{}^{\circ }$ , and use them to find the unknown lengths of the right-angled triangles.

.g. $\sin 30{}^{\circ }={\displaystyle \frac{1}{2}},\cos 30{}^{\circ }={\displaystyle \frac{\sqrt{3}}{2}},\tan 30{}^{\circ }={\displaystyle \frac{1}{\sqrt{3}}}$
$\sin 60{}^{\circ }={\displaystyle \frac{\sqrt{3}}{2}},\cos 60{}^{\circ }={\displaystyle \frac{1}{2}},\tan 60{}^{\circ }=\sqrt{3}$

Complete step by step solution:
Let's say that AB is the cliff and CD is the tower, as shown in the following diagram:

Using the definition of angle of depression and by the properties of parallel lines, we have $\mathrm{\angle }ACB=30{}^{\circ }$ and $\mathrm{\angle }ADB=60{}^{\circ }$ . Also, $AB=200m$ (height of the cliff).

Using the definition of $\tan \theta$ in $\mathrm{\Delta }ABD$ , we have:
$\tan 60{}^{\circ }={\displaystyle \frac{P}{B}}={\displaystyle \frac{AB}{BD}}={\displaystyle \frac{200}{BD}}$

Using $\tan 60{}^{\circ }=\sqrt{3}$ , we get:
⇒ $\sqrt{3}={\displaystyle \frac{200}{BD}}$
⇒ $BD={\displaystyle \frac{200}{\sqrt{3}}}=115.47m$

Now $CX=BD={\displaystyle \frac{200}{\sqrt{3}}}$ (why?). Using the definition of $\tan \theta$ in $\mathrm{\Delta }AXC$ ,

we have:
$\tan 30{}^{\circ }={\displaystyle \frac{P}{B}}={\displaystyle \frac{AX}{XC}}$

Using $\tan 30{}^{\circ }={\displaystyle \frac{1}{\sqrt{3}}}$ , we get:
⇒ $AX={\displaystyle \frac{1}{\sqrt{3}}}XC={\displaystyle \frac{1}{\sqrt{3}}}\times {\displaystyle \frac{200}{\sqrt{3}}}={\displaystyle \frac{200}{3}}=66{\displaystyle \frac{2}{3}}m$
And $CD=XB=AB-AX$ .
⇒ $CD=200-66{\displaystyle \frac{2}{3}}=133{\displaystyle \frac{1}{3}}m$

The correct answer is B. Height = $133{\displaystyle \frac{1}{3}}$ m; Distance = 115.46 m.

Note:
In a right-angled triangle with length of the side opposite to angle θ as perpendicular (P), base (B) and hypotenuse (H):
$\sin \theta ={\displaystyle \frac{P}{H}},\cos \theta ={\displaystyle \frac{B}{H}},\tan \theta ={\displaystyle \frac{P}{B}}$
$P^2+B^2=H^2$ (Pythagoras' Theorem)

If one trigonometric ratio is known, we can use Pythagoras' Theorem and calculate the values of all other trigonometric ratios.