Question

Solve this fast plz........!!!!​

Answer 1

Given:

• Angle of elevation of the top of a tower from two points distant s and t from its foot are complementary.

Prove That:

• Height of the tower is √(st)

Solution:

Let, Height of the tower, CD = h metre

Let, A and B any two points on the ground such that they make a distant s and t from the foot of the tower.

At points A and B, they make an angle of elevation to the top of the tower are

✒ $\sf \angle CAD = \theta \: and \: \angle CBD = {90}^{ \circ} - \theta$

($\because$ In complementary the sum of two angles are 90°)

☃ $\sf In \: right \: angled \: \triangle CAD$

$\to\sf \tan \theta = \dfrac{CD}{AC}$

$\implies\sf \tan \theta = \dfrac{h}{s} \: .......(i)$

☃ $\sf And \: in \: right \: angled \: \triangle CBD,$

$\sf \to \tan ( {90}^{ \circ} - \theta )= \dfrac{CD}{BC}$

$\implies\sf \cot \theta = \dfrac{h}{t} \: .......(ii)$

✏$\pink{On \: multiplying \: eq.(i) \: and \: eq.(ii), \: we \: get}$

$\sf \hookrightarrow \tan \theta \times \cot \theta = \dfrac{h}{s} \times \dfrac{h}{t}$

$\sf \implies 1 = \dfrac{ {h}^{2} }{st}$

$\sf \implies {h}^{2} = st$

$\sf \implies h= \sqrt{st}$

Hence, Proved