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MHTCET 22 Apr 2025 Shift 1 Paper

Question : 108 of 150

Marks: +1, -0

\triangle A B C

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

\angle C=\ldots

Solution:

Concept:

Use the cosine rule and square the derived relation to connect the given condition with

\cos C

Explanation:

Given equation:

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

\Rightarrow a^{4}+b^{4}+c^{4}=2a^{2}c^{2}+2b^{2}c^{2}

From the cosine rule,

\cos C=\frac{a^{2}+b^{2}-c^{2}}{2ab}

\Rightarrow a^{2}+b^{2}-c^{2}=2ab\cos C

Squaring both sides,

(a^{2}+b^{2}-c^{2})^{2}=4a^{2}b^{2}\cos^{2}C

Expanding the left side,

a^{4}+b^{4}+c^{4}+2a^{2}b^{2}-2a^{2}c^{2}-2b^{2}c^{2}=4a^{2}b^{2}\cos^{2}C

Substituting

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

2a^{2}c^{2}+2b^{2}c^{2}+2a^{2}b^{2}-2a^{2}c^{2}-2b^{2}c^{2}=4a^{2}b^{2}\cos^{2}C

2a^{2}b^{2}=4a^{2}b^{2}\cos^{2}C

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

\cos C=\pm\frac{1}{\sqrt{2}}

C=45^{\circ}

135^{\circ}

Since the given options include only

135^{\circ}

, the intended value is

135^{\circ}

Answer:

\angle C=135^{\circ}

Option A:

135^{\circ}

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