Question

triangle abc angle b is equal to 90 degree angle C is equal to 60 degree AC is equal to 10 cm find the other two sides

Answer 1

you can do it in both ways
Hope it helps

Answer 2

$In \: \triangle ABC, \angle B = 90\degree$

$\angle C= 60 \degree \: and \: AC = 10 \:cm$

$i ) sin \:60\degree= \frac{AB}{AC}$

$\implies \frac{\sqrt{3}}{2} = \frac{AB}{10}$

$\implies 10 \times \frac{\sqrt{3}}{2} = AB$

$\therefore AB = 5\sqrt{3} \:cm$

$ii ) cos \:60\degree= \frac{BC}{AC}$

$\implies \frac{1}{2} = \frac{BC}{10}$

$\implies 10 \times \frac{1}{2} = BC$

$\therefore BC = 5 \:cm$

$\red { Required \: other \:two \: sides \: are }\\\green { \: AB = 5\sqrt{3} \:cm \:and \: BC = 5 \:cm }$

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