Prove that : a^2sin2c + c^2 sin2A = 4∆ By using tangent rule and half angels rule
Prove that : a^2sin2c + c^2 sin2A = 4∆
By using tangent rule and half angels rule
Sampath sriram , 7 Years ago
Grade 11
Magical Mathematics[Interesting Approach]> Prove that : a^2sin2c + c^2 sin2A = 4∆ By...
1 Answers
Arun
Last Activity: 7 Years ago
b2sin2C + c2sin2B = b2(2sinC cosC) + c2(2sinB cosB) Since, sinC/c =sinB/b = sinA/a = R So, = b2(2Rc cosC) + c2(2RbcosB) =b2[2Rc ×{b2+a2 - c2}/(2ab)]+c2[2Rb×{a2 + c2-b2}/(2ac)] =2Rbc [ {b2+a2 - c2}/(2a) + {a2 + c2-b2}/(2a)] =2Rbc [ a] = 2bc(Ra) =2bcsinA = 4 * Area of triangle ABC
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