In a Δ ABC, a^{4}+b^{4}+c^{4}=2b²c²+2a²b² then B=

Correct Answer is : π4π/44π or 3π43π/443π

Correct Answer is : π4π/44π or 3π43π/443π

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TS EAMCET 4 May 2018 Shift 2 Solved Paper

Section: Mathematics

Question : 28 of 160

Marks: +1, -0

In a

\Delta ABC, a^{4}+b^{4}+c^{4}=2b^{2}c^{2}+2a^{2}b^{2}

B=

Solution:

Consider the expression,

a^{4}+b^{4}+c^{4}=2b^{2}(c^{2}+a^{2})

a^{4}+b^{4}+c^{4}-2b^{2}c^{2}-2a^{2}b^{2}=0

Add

2a^{2}c^{2}

on both sides.

a^{4}+b^{4}+c^{4}-2b^{2}c^{2}-2a^{2}b^{2}

+2c^{2}a^{2}=2c^{2}a^{2}

(a^{2}+c^{2}-b^{2})^{2}=2c^{2}a^{2}

\frac{4(a^{2}+c^{2}-b^{2})^{2}}{(2ca)^{2}}=2

\frac{(a^{2}+c^{2}-b^{2})^{2}}{(2ca)^{2}}=\frac{1}{2}

This implies,

\cos^{2}B=\frac{1}{2}

\cos B=\frac{1}{\sqrt{2}}

So,

angle

B=

angle

C=\frac{\pi}{4}

\frac{3\pi}{4}

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