Question

The angles of elevation of the top of a tower from two points at a distance of 4m and 9m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6m. ​

Answer 1

Given :-

The angles of elevation of the top of a tower from two points at a distance of 4m and 9m from the base of the tower and in the same straight line with it are complementary.

To prove :-

• ★ the height of the tower is 6m. 

$\huge\bf\underline \red{Solution:-}$

• AC = 9m
• AD = 4m

Let ∠ACB = ∅ and , ∠BDA = (90° - ∅)

Let the height of the tower be h meter.

Now,

In triangle ∆ ABC

$\bf \pink{ \tan \theta = \frac{perpendicular}{base} }$

$\rm \longmapsto \: tan \theta \: = \frac{AB}{AC} \\ \\ \rm \longmapsto \: tan \theta \: = \frac{h}{9} ................(i)$

Now,

In triangle ∆ ABD,

$\pink{\bf\longmapsto \: tan \:( 90 \degree - \theta )\: = cot \theta}$

$\rm \longmapsto \: tan (90 \degree - \theta )\: = \frac{AB}{AD} \\ \\ \rm \longmapsto \: tan (90 \degree - \theta )\: = \frac{h}{4} \\ \\ \rm \longmapsto \: cot \theta = \frac{h}{4} ............(ii)\:$

• ★ Multiplying both the equations:-

$\rm \longmapsto \: tan \theta \times \: cot \theta = \frac{h}{9} \times \frac{h}{4} \\ \\ \bf \longmapsto \pink{tan \theta \times \: cot \theta = 1} \\ \\ \rm \longmapsto \:1= \frac{ {h}^{2} }{36} \\ \\\rm \longmapsto \: h = \sqrt{36}\\ \\ \bf \longmapsto \: h =\pm\: 6m$

height can not be negative .

So, the height of the tower is 6m.

Hence Proved that the height of the tower is 6m.