Question
Find $AB.$
✓
Answer
Consider the figure
From right $\triangle ACF$
$\tan 45^{\circ}=\frac{20}{ AC }$
$1=\frac{20}{ AC }$
$AC =20 \ cm$
From $\triangle DEB$
$\tan 60^{\circ}=\frac{30}{ BD }$
$\sqrt{3}=\frac{30}{ BD }$
$BD =\frac{30}{\sqrt{3}}=17.32 \ cm$
Given $FC = 20, ED = 30,$ So $EP = 10 \ cm$
Therefore
$\tan 60^{\circ}=\frac{ FP }{ EP }$
$\sqrt{3}=\frac{ FP }{10}$
$FP =10 \sqrt{3}=17.32 \ cm $
Thus $AB = AC + CD + BD = 54.64 \ cm$
From right $\triangle ACF$
$\tan 45^{\circ}=\frac{20}{ AC }$
$1=\frac{20}{ AC }$
$AC =20 \ cm$
From $\triangle DEB$
$\tan 60^{\circ}=\frac{30}{ BD }$
$\sqrt{3}=\frac{30}{ BD }$
$BD =\frac{30}{\sqrt{3}}=17.32 \ cm$
Given $FC = 20, ED = 30,$ So $EP = 10 \ cm$
Therefore
$\tan 60^{\circ}=\frac{ FP }{ EP }$
$\sqrt{3}=\frac{ FP }{10}$
$FP =10 \sqrt{3}=17.32 \ cm $
Thus $AB = AC + CD + BD = 54.64 \ cm$
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