A tower PQ stands on a horizontal ground with base Q on the ground. The point R divides the tower in two parts such that QR =15 m. If from a point A on the ground the angle of elevation of R is 60° and the part PR of the tower subtends an angle of 15° at A, then the height of the tower is :

Question

A tower PQ stands on a horizontal ground with base Q on the ground. The point R divides the tower in two parts such that QR =15 m. If from a point A on the ground the angle of elevation of R is 60° and the part PR of the tower subtends an angle of 15° at A, then the height of the tower is : (A) 5(2 √3+3) m (B) 5(√3+3) m (C) 10(√3+1) m (D) 10(2 √3+1) m

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/101cebb21956e34cb46044298f7d0d3c-.png" /> (15/ AQ )= tan 60° (15+x/ AQ )= tan 75° ((1)/(2)) ⇒ x=10 √3So, P Q=5(2 √3+3) m

$5 (23 + 3) m$

$5 (3 + 3) m$

$10 (3 + 1) m$

$10 (23 + 1) m$

$\frac{15}{A Q} = tan 6 0^{\circ}$
$\frac{15 + x}{A Q} = tan 7 5^{\circ}$
$\frac{( 1 )}{( 2 )} \Rightarrow x = 103$ So, $PQ = 5 (23 + 3) m$

Q. A tower PQ stands on a horizontal ground with base Q on the ground. The point R divides the tower in two parts such that QR=15m. If from a point A on the ground the angle of elevation of R is 60∘ and the part PR of the tower subtends an angle of 15∘ at A, then the height of the tower is :

5(23​+3)m

5(3​+3)m

10(3​+1)m

10(23​+1)m

AQ15​=tan60∘ AQ15+x​=tan75∘ (2)(1)​⇒x=103​So, PQ=5(23​+3)m

$5 (23 + 3) m$

$5 (3 + 3) m$

$10 (3 + 1) m$

$10 (23 + 1) m$

$\frac{15}{A Q} = tan 6 0^{\circ}$
$\frac{15 + x}{A Q} = tan 7 5^{\circ}$
$\frac{( 1 )}{( 2 )} \Rightarrow x = 103$ So, $PQ = 5 (23 + 3) m$