Question
In $\triangle ABC, AB = AC = 15 \ cm$ and $BC = 18 \ cm,$ find $\cos \angle ABC.$
✓
Answer
Consider the diagram below :
In the isosceles $\triangle ABC, AB = AC = 15\ cm$ and $BC =18\ cm$ the perpendicular drawn from angle $A$ to the side $BC$ divides the side $BC$ into two equal parts $BD = DC = 9 \ cm$
$\cos \angle ABC =\frac{\text { base }}{\text { hypotenuse }}=\frac{ BD }{ AB }=\frac{9}{15}=\frac{3}{5}$
In the isosceles $\triangle ABC, AB = AC = 15\ cm$ and $BC =18\ cm$ the perpendicular drawn from angle $A$ to the side $BC$ divides the side $BC$ into two equal parts $BD = DC = 9 \ cm$
$\cos \angle ABC =\frac{\text { base }}{\text { hypotenuse }}=\frac{ BD }{ AB }=\frac{9}{15}=\frac{3}{5}$
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