TS EAMCET · Maths · Properties of Triangles
In a triangle
- A
- B
- C
- D
Answer & Solution
Correct Answer
(B)
Step-by-step Solution
Detailed explanation
b2sin2C+c2sin2B =4R2sin2B2sinCcosC+4R2sin2c2sinBcosB ∵asinA=bsinB=csinC=2R =8R2sinBsinCsinBcosC+cosBsinC =8R2sinBsinCsinB+C =8R2sinBsinCsinπ-A =8R2sinBsinCsinA =22RsinB2RsinCsinA, =2bcsinA {Using sine rule} =412bcsinA=4∆
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