Question
Find $AB$ and $BC,$ if:
✓
Answer
Let $BC = x \ m$
$BD = BC + CD = (x + 20) \ cm$
In $\triangle ABD,$
$\tan 30^{\circ}=\frac{ AB }{ BD }$
$\frac{1}{\sqrt{3}}=\frac{ AB }{x+20}$
$x+20=\sqrt{3} A B\dots ...(1)$
In $\triangle ABC$
$\tan 45^{\circ}=\frac{ AB }{ BC }$
$1=\frac{ AB }{x}$
$AB = x\dots ...(2)$
From $(1)$
$A B+20=\sqrt{3} A B$
$AB (\sqrt{3}-1)=20$
$ AB =\frac{20}{(\sqrt{3}-1)}$
$=\frac{20}{(\sqrt{3}-1)} \times \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}$
$=\frac{20(\sqrt{3}+1)}{3-1}$
$= 27.32\ cm$
From $(2)$
$AB = x = 27.32 \ cm$
Therefore $BC = x = AB = 27.32 \ cm$
Therefore $, AB = 27.32 \ cm, BC = 27.32 \ cm$
$BD = BC + CD = (x + 20) \ cm$
In $\triangle ABD,$
$\tan 30^{\circ}=\frac{ AB }{ BD }$
$\frac{1}{\sqrt{3}}=\frac{ AB }{x+20}$
$x+20=\sqrt{3} A B\dots ...(1)$
In $\triangle ABC$
$\tan 45^{\circ}=\frac{ AB }{ BC }$
$1=\frac{ AB }{x}$
$AB = x\dots ...(2)$
From $(1)$
$A B+20=\sqrt{3} A B$
$AB (\sqrt{3}-1)=20$
$ AB =\frac{20}{(\sqrt{3}-1)}$
$=\frac{20}{(\sqrt{3}-1)} \times \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}$
$=\frac{20(\sqrt{3}+1)}{3-1}$
$= 27.32\ cm$
From $(2)$
$AB = x = 27.32 \ cm$
Therefore $BC = x = AB = 27.32 \ cm$
Therefore $, AB = 27.32 \ cm, BC = 27.32 \ cm$
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