Question 514 Marks
A ladder rests against a tree on one side of a street. The foot of the ladder makes an angle of 50° with the ground. When the ladder is turned over to rest against another tree on the other side of the street it makes an angle of 40° with the ground. If the length of the ladder is 60m, find the width of the street.
Answer
Let $A B$ and $C D$ be two trees and $P$ be a point on the street $A C$ between the two trees.
PD and $\mathrm{PB}$ denotes the ladder at the two instants.
$\ln \triangle \mathrm{PCD}$
$
\begin{aligned}
& \cos 50^{\circ}=\frac{P C}{P D} \\
& 0.6428=\frac{P C}{60} \\
& \Rightarrow P C=0.6428 \times 60=38.568
\end{aligned}
$
$\ln \triangle \mathrm{ABP}$
$\cos 40^{\circ}=\frac{\mathrm{AP}}{\mathrm{BP}}$
$
\Rightarrow 0.7660=\frac{\mathrm{AP}}{60}
$
$
\Rightarrow \mathrm{AP}=0.7660 \times 60=45.96
$
$
\therefore A C=A P+P C=38.568 \mathrm{~m}+45.96 \mathrm{~m}=84.528 \mathrm{~m} \approx 84.53 \mathrm{~m} .
$
Thus, the width of the street is $84.53 \mathrm{~m}$.
View full question & answer→Let $A B$ and $C D$ be two trees and $P$ be a point on the street $A C$ between the two trees.
PD and $\mathrm{PB}$ denotes the ladder at the two instants.
$\ln \triangle \mathrm{PCD}$
$
\begin{aligned}
& \cos 50^{\circ}=\frac{P C}{P D} \\
& 0.6428=\frac{P C}{60} \\
& \Rightarrow P C=0.6428 \times 60=38.568
\end{aligned}
$
$\ln \triangle \mathrm{ABP}$
$\cos 40^{\circ}=\frac{\mathrm{AP}}{\mathrm{BP}}$
$
\Rightarrow 0.7660=\frac{\mathrm{AP}}{60}
$
$
\Rightarrow \mathrm{AP}=0.7660 \times 60=45.96
$
$
\therefore A C=A P+P C=38.568 \mathrm{~m}+45.96 \mathrm{~m}=84.528 \mathrm{~m} \approx 84.53 \mathrm{~m} .
$
Thus, the width of the street is $84.53 \mathrm{~m}$.
