Question
Find $PQ,$ if $AB = 150\ m, \angle P = 30^\circ$ and $\angle Q = 45^\circ.$
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Answer
From $\triangle APB$
$\tan 30^{\circ}=\frac{ AB }{ PB } $
$\frac{1}{\sqrt{3}}=\frac{150}{ PB }$
$PB =150 \sqrt{3}$
$PB = 259.80\ m$
Also, from $\triangle ABQ$
$\tan 45^{\circ}=\frac{ AB }{ BQ }$
$1=\frac{150}{ BQ }$
$BQ = 150\ m$
Therefore,
$PQ = PB - BQ$
$PQ = 259.80 - 150$
$PQ = 109.80\ m$
$\tan 30^{\circ}=\frac{ AB }{ PB } $
$\frac{1}{\sqrt{3}}=\frac{150}{ PB }$
$PB =150 \sqrt{3}$
$PB = 259.80\ m$
Also, from $\triangle ABQ$
$\tan 45^{\circ}=\frac{ AB }{ BQ }$
$1=\frac{150}{ BQ }$
$BQ = 150\ m$
Therefore,
$PQ = PB - BQ$
$PQ = 259.80 - 150$
$PQ = 109.80\ m$
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