Two towers A and B have heights 45 , m and 15 | vedclass

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(A) Let the height of tower $A$ be $AP = 45 \, m$ and the height of tower $B$ be $BQ = 15 \, m$. Let the distance between the bases of the towers be $

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) Let the height of tower $A$ be $AP = 45 \, m$ and the height of tower $B$ be $BQ = 15 \, m$. Let the distance between the bases of the towers be $RO = d$.In $\Delta PRO$,the angle of elevation from the bottom of tower $B$ $(R)$ to the top of tower $A$ $(P)$ is $60^{\circ}$.$	an 60^{\circ} = \frac{AP}{RO} = \frac{45}{d}$$\sqrt{3} = \frac{45}{d} \implies d = \frac{45}{\sqrt{3}} = 15\sqrt{3} \, m$.Now,in $\Delta QOR$,the angle of elevation from the bottom of tower $A$ $(O)$ to the top of tower $B$ $(Q)$ is $	heta$.$	an 	heta = \frac{BQ}{RO} = \frac{15}{15\sqrt{3}} = \frac{1}{\sqrt{3}}$.Since $	an 30^{\circ} = \frac{1}{\sqrt{3}}$,we have $	heta = 30^{\circ}$.Therefore,$\sin 	heta = \sin 30^{\circ} = \frac{1}{2}$.

Two towers $A$ and $B$ have heights $45 \, m$ and $15 \, m$ respectively. The angle of elevation from the bottom of tower $B$ to the top of tower $A$ is $60^{\circ}$. If the angle of elevation from the bottom of tower $A$ to the top of tower $B$ is $\theta$,then the value of $\sin \theta$ is:

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