The angle of elevation of building Y from the | vedclass

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(C) Let $X$ be the height of building $PQ$ and $Y$ be the height of building $SR$. Let $QR$ be the distance between the bases of the two buildings.In

Vedclass · Accepted Answer

Height of $X$ is greater than $Y$

Vedclass · Answer

(C) Let $X$ be the height of building $PQ$ and $Y$ be the height of building $SR$. Let $QR$ be the distance between the bases of the two buildings.In right-angled $\Delta SQR$,$	an(45^{\circ}) = \frac{SR}{QR} = \frac{Y}{QR}$.Since $	an(45^{\circ}) = 1$,we have $Y = QR$.In right-angled $\Delta PQR$,$	an(65^{\circ}) = \frac{PQ}{QR} = \frac{X}{QR}$.Since $	an(65^{\circ}) > 1$,we have $\frac{X}{QR} > 1$,which implies $X > QR$.Since $Y = QR$ and $X > QR$,it follows that $X > Y$.Therefore,the height of building $X$ is greater than the height of building $Y$.

The angle of elevation of building $Y$ from the bottom of building $X$ is $45^{\circ}$ and the angle of elevation of building $X$ from the bottom of building $Y$ is $65^{\circ}.$ Then:

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