Question
Find $AB$ and $BC,$ if:
✓
Answer
Let $BC = x \ m$
$BD = BC + CD = (x + 20) \ cm$
In $\triangle ABD,$
$\tan 45^\circ =\frac{A B}{B D}$
$1=\frac{ AB }{x+20}$
$x + 20 = AB \dots...(1)$
In $\triangle ABC$
$\tan 60^{\circ}=\frac{ AB }{ BC }$
$\sqrt{3}=\frac{ AB }{x}$
$x =\frac{ AB }{\sqrt{3}} \dots...(2)$
From $(1)$
$\frac{ AB }{\sqrt{3}}+20= AB$
$AB +20 \sqrt{3}=\sqrt{3} AB$
$AB (\sqrt{3}-1)=20 \sqrt{3}$
$ AB =\frac{20 \sqrt{3}}{(\sqrt{3}-1)}$
$AB =\frac{20 \sqrt{3}}{(\sqrt{3}-1)} \times \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}$
$AB =\frac{20 \sqrt{3}(\sqrt{3}+1)}{3-1}$
$AB = 47.32 \ cm$
From $(2)$
$x=\frac{A B}{\sqrt{3}}$
$x=\frac{47.32}{\sqrt{3}}$
$x = 27.32 \ cm$
$\therefore BC = x = 27.32 \ cm$
Therefore $, AB = 47.32 \ cm, BC = 27.32 \ cm.$
$BD = BC + CD = (x + 20) \ cm$
In $\triangle ABD,$
$\tan 45^\circ =\frac{A B}{B D}$
$1=\frac{ AB }{x+20}$
$x + 20 = AB \dots...(1)$
In $\triangle ABC$
$\tan 60^{\circ}=\frac{ AB }{ BC }$
$\sqrt{3}=\frac{ AB }{x}$
$x =\frac{ AB }{\sqrt{3}} \dots...(2)$
From $(1)$
$\frac{ AB }{\sqrt{3}}+20= AB$
$AB +20 \sqrt{3}=\sqrt{3} AB$
$AB (\sqrt{3}-1)=20 \sqrt{3}$
$ AB =\frac{20 \sqrt{3}}{(\sqrt{3}-1)}$
$AB =\frac{20 \sqrt{3}}{(\sqrt{3}-1)} \times \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}$
$AB =\frac{20 \sqrt{3}(\sqrt{3}+1)}{3-1}$
$AB = 47.32 \ cm$
From $(2)$
$x=\frac{A B}{\sqrt{3}}$
$x=\frac{47.32}{\sqrt{3}}$
$x = 27.32 \ cm$
$\therefore BC = x = 27.32 \ cm$
Therefore $, AB = 47.32 \ cm, BC = 27.32 \ cm.$
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