From the base of a pole of height 20 meter, the angle of elevation of the top of a tower is 60°. The pole subtends an angle 30° at the top of the tower. Then the height of the tower is:

Question

From the base of a pole of height 20 meter, the angle of elevation of the top of a tower is 60°. The pole subtends an angle 30° at the top of the tower. Then the height of the tower is: (A) 15 √3 (B) 20 √3 (C) 20+10 √3 (D) 30

Tardigrade Admin · Accepted Answer

PT =( h /√3)= AB ( AB / h -20)=√3 h =3( h -20) h =30 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/07eb1a99e7a3a07f7add521c934caa37-.png" />

Q. From the base of a pole of height $20$ meter, the angle of elevation of the top of a tower is $6 0^{\circ}$ . The pole subtends an angle $3 0^{\circ}$ at the top of the tower. Then the height of the tower is:

$153$

$203$

$20 + 103$

$30$

$PT = \frac{h}{3} = A B$
$\frac{A B}{h - 20} = 3$
$h = 3 (h - 20)$
$h = 30$

Q. From the base of a pole of height 20 meter, the angle of elevation of the top of a tower is 60∘. The pole subtends an angle 30∘ at the top of the tower. Then the height of the tower is:

153​

203​

20+103​

30