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Question : 26 of 160
Marks:
+1,
-0
Solution:
We have,
ABC is a triangle
‌cos‌A+cos‌B+cos‌C‌=2‌cos(‌)‌cos(‌)+1−2sin‌2‌‌=1+2sin‌‌‌cos(‌)−2sin‌2‌‌=1+2sin‌‌[cos(‌)−sin‌‌]‌=1+2sin‌‌[cos(‌)−cos(‌)]‌=1+2sin‌‌[2sin‌‌sin‌‌]‌=1+4sin‌‌sin‌‌sin‌‌‌=1+4[√‌√‌]‌=‌| 1+4(s−a)(s−b)(s−c) |
| abc |
‌=‌| 1+4s(s−a)(s−b)(s−c) |
| sabc |
[∴‌ multiply and divide by ‌s]‌=1+4‌,‌ where ‌∆‌ is area of triangle ‌‌=1+(‌)(‌)‌‌[∵r=‌‌ and ‌R=‌]‌=1+‌
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