The angle of elevation of the top of a TV tow | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $OP = h$ be the height of the tower,where $O$ is the foot of the tower.In $	riangle OAP$,$\angle OAP = \alpha$. Thus,$OA = h \cot \alpha$.In

Vedclass · Accepted Answer

$a \sin 2 \alpha$

Vedclass · Answer

(C) Let $OP = h$ be the height of the tower,where $O$ is the foot of the tower.In $	riangle OAP$,$\angle OAP = \alpha$. Thus,$OA = h \cot \alpha$.In $	riangle OBP$,$\angle OBP = 2 \alpha$. Thus,$OB = h \cot 2 \alpha$.Given $AB = a$,we have $OA - OB = a$.Therefore,$h \cot \alpha - h \cot 2 \alpha = a$.$h (\frac{\cos \alpha}{\sin \alpha} - \frac{\cos 2 \alpha}{\sin 2 \alpha}) = a$.$h (\frac{\sin 2 \alpha \cos \alpha - \cos 2 \alpha \sin \alpha}{\sin \alpha \sin 2 \alpha}) = a$.Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$,we get:$h (\frac{\sin(2 \alpha - \alpha)}{\sin \alpha \sin 2 \alpha}) = a$.$h (\frac{\sin \alpha}{\sin \alpha \sin 2 \alpha}) = a$.$h (\frac{1}{\sin 2 \alpha}) = a$.$h = a \sin 2 \alpha$.Thus,the height of the tower is $a \sin 2 \alpha$.

The angle of elevation of the top of a $TV$ tower from three points $A, B, C$ in a straight line through the foot of the tower are $\alpha, 2 \alpha, 3 \alpha$ respectively. If $AB = a$,the height of the tower is

Similar Questions

The angle of elevation of a tower from a distance of $100 \ m$ from its foot is $30^{\circ}$. Then the height of the tower is (in $m$):

$A$ tower standing on a horizontal plane subtends a certain angle at a point $160 \; m$ apart from the foot of the tower. On advancing $100 \; m$ towards it,the tower is found to subtend an angle twice as before. The height of the tower is (in $m$):

$A$ tower stands on a horizontal plane. $A$ man on the ground $100 \ m$ from the base of the tower finds the angle of elevation of the top of the tower to be $30^{\circ}$. What is the height of the tower (in $m$)?

The angle of elevation of the top of a tower at a point $G$ on the ground is $30^{\circ}$. On walking $20 \ m$ towards the tower,the angle of elevation becomes $60^{\circ}$. The height of the tower is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module