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Question : 198 of 200
Marks:
+1,
-0
Solution:
We have,
‌‌+‌=0‌⇒‌‌‌| 2Rsin‌A |
| 4R2(sin‌2B−sin‌2C) |
‌‌‌+‌| 2Rsin‌C |
| 4R2(sin‌2B−sin‌2A) |
=0 ‌⇒‌‌‌⋅‌| sin‌A |
| sin‌(B+C)sin‌(B−C) |
‌+‌⋅‌| sin‌C |
| sin‌(B+A)sin‌(B−A) |
=0‌⇒‌‌‌⋅‌| sin‌A |
| sin‌A⋅sin‌(B−C) |
‌+‌⋅‌| sin‌C |
| sin‌Csin‌(B−A) |
=0‌⇒‌‌‌+‌=0‌⇒‌‌sin‌(B−C)+sin‌(B−A)=0‌⇒‌‌2sin‌‌‌cos‌=0‌⇒‌‌sin‌‌=0‌⇒‌‌2B=A+C‌⇒‌‌2B=180∘−B‌⇒‌‌3B=180∘‌⇒‌‌B=‌‌
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