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Question : 26 of 160
Marks:
+1,
-0
Solution:
In
â–³ABC,
‌∠A=60∘,∠B+∠C=180∘−∠A=120∘ ;a=k‌sin‌A,b=k‌sin‌B,c=k‌sin‌C‌ (sine rule) ‌ Now,(b+c+a)(b+c−a) ‌=(b+c)2−a2 ‌=k2(sin‌B+sin‌C)2−k2‌sin‌2A ‌=K2(2sin‌‌‌cos‌)2−K2(‌) ‌[∵sin‌A=‌] ‌=k2(2‌sin‌60∘⋅cos‌)2−‌ ‌=3k2cos2‌−‌ ‌=3k2(cos2‌−‌) ‌=‌(2cos2‌−1+‌) ‌=‌(cos(B−C)+‌) ‌=‌(cos(B−C)+cos‌A) ‌=‌(cos(B−C)−cos(B+C)) ‌=‌×2‌sin‌B×sin‌C ‌=3(k‌sin‌B)(k‌sin‌C)=3bc ‌
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