Question

- (0) In the triangle ABC, a=4, b=8, c=60º find angle B. त्रिभुज ABC में कोण B का मान कीजिए जहां a=4, b=8, c=60°

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Solution

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( =2.5 x ) ( k ) ( frac{sqrt{8}}{60} c_{4}=2 times 10 ) ( sqrt{3} times frac{1}{4} ) ( mathcal{V} ) ( c^{2}=a^{2}+b^{2}-2 a b cos c ) ( begin{aligned}=& 64+16-64left(frac{1}{2}right) =& 48 C=sqrt{48}=4 sqrt{3} frac{4 sqrt{3}}{sin (60)}=frac{8}{8 i n(13)} end{aligned} ) 7) ( sin B=frac{8 x sqrt{3} / 2}{4 sqrt{3}}=1 quad therefore quad B=90^{circ} )