A tower is 30 m high. An observer from the t | vedclass

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

Question

(D) Let the height of the building $AB$ be $H$ metres.Let $Q$ be the top of the tower and $P$ be the base of the tower. Given $QP = 30 \ m$.In $\Delta

Vedclass · Accepted Answer

$12.68$

Vedclass · Answer

(D) Let the height of the building $AB$ be $H$ metres.Let $Q$ be the top of the tower and $P$ be the base of the tower. Given $QP = 30 \ m$.In $\Delta APQ$,the angle of elevation from $A$ to $Q$ is equal to the angle of depression from $Q$ to $A$,which is $60^{\circ}$.$	an 60^{\circ} = \frac{QP}{AP} = \frac{30}{AP} = \sqrt{3}$$AP = \frac{30}{\sqrt{3}} = 10 \sqrt{3} \ m$.Let $R$ be a point on $QP$ such that $BR$ is horizontal. Then $BR = AP = 10 \sqrt{3} \ m$.The angle of elevation from $B$ to $Q$ is equal to the angle of depression from $Q$ to $B$,which is $45^{\circ}$.In $\Delta BRQ$,$	an 45^{\circ} = \frac{QR}{BR} = 1$.$QR = BR = 10 \sqrt{3} \ m$.The height of the building $H = AB = RP = QP - QR$.$H = 30 - 10 \sqrt{3} = 30 - 10(1.732) = 30 - 17.32 = 12.68 \ m$.

$A$ tower is $30 \ m$ high. An observer from the top of the tower makes an angle of depression of $60^{\circ}$ at the base of the building and an angle of depression of $45^{\circ}$ at the top of the building. What is the height of the building in metres?

Similar Questions

At a point,$15 \, m$ away from the base of a $15 \, m$ high house,the angle of elevation of the top is (in $^{\circ}$)

The length of the shadow of a tower is $9 \text{ m}$ when the sun's altitude is $30^{\circ}$. What is the height of the tower? (In $m$)

Two towers of equal height stand on either side of a wide road which is $100 \, m$ wide. At a point on the road between the towers, the angles of elevation of the tops of the towers are $60^{\circ}$ and $30^{\circ}$. Find their heights in metres. (in $\sqrt{3}$)

The tops of two poles of height $60 \ m$ and $35 \ m$ are connected by a rope. If the rope makes an angle with the horizontal whose tangent is $\frac{5}{9}$,then what is the distance (in $m$) between the two poles?

From a point $20 \; m$ away from the foot of a tower,the angle of elevation of the top of the tower is $30^{\circ}$. The height of the tower is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module