Question
Find $AB$ and $BC$, if:
✓
Answer
Let $BC = x m$
$BD = BC + CD = (x + 20) \ cm$
In $\triangle ABD,a$
$\tan 30^\circ = \frac{ AB }{ BD }$
$\frac{1}{\sqrt{3}}=\frac{ AB }{x+20}$
$x+20=\sqrt{3} A B \dots...(1)$
In $\triangle ABC$
$\tan 60^{\circ}=\frac{ AB }{ BC }$
$\sqrt{3}=\frac{ AB }{x}$
$x =\frac{ AB }{\sqrt{3}} \dots...(2)$
From $(1)$
$\frac{ AB }{\sqrt{3}}+20=\sqrt{3} AB$
$AB +20 \sqrt{3}=3 AB$
$2 AB =20 \sqrt{3}$
$2 AB =\frac{20 \sqrt{3}}{2}$
$AB =10 \sqrt{3}$
$AB =17.32 \ cm $
From $(2)$
$x=\frac{A B}{\sqrt{3}}$
$x=\frac{17.32}{\sqrt{3}}$
$x = 10 \ cm$
Therefore $BC = x = 10 \ cm$
Therefore, $AB = 17.32 \ cm, BC = 10 \ cm.$
$BD = BC + CD = (x + 20) \ cm$
In $\triangle ABD,a$
$\tan 30^\circ = \frac{ AB }{ BD }$
$\frac{1}{\sqrt{3}}=\frac{ AB }{x+20}$
$x+20=\sqrt{3} A B \dots...(1)$
In $\triangle ABC$
$\tan 60^{\circ}=\frac{ AB }{ BC }$
$\sqrt{3}=\frac{ AB }{x}$
$x =\frac{ AB }{\sqrt{3}} \dots...(2)$
From $(1)$
$\frac{ AB }{\sqrt{3}}+20=\sqrt{3} AB$
$AB +20 \sqrt{3}=3 AB$
$2 AB =20 \sqrt{3}$
$2 AB =\frac{20 \sqrt{3}}{2}$
$AB =10 \sqrt{3}$
$AB =17.32 \ cm $
From $(2)$
$x=\frac{A B}{\sqrt{3}}$
$x=\frac{17.32}{\sqrt{3}}$
$x = 10 \ cm$
Therefore $BC = x = 10 \ cm$
Therefore, $AB = 17.32 \ cm, BC = 10 \ cm.$
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