In a triangle ABC, BC=7, AC=8, AB=α ∈ N and A=(2)/(3) . If 49 (3 C)+42= m n , where gcd(m, n)=1 , then m+n is equal to________

Correct answer: 0 aligned & A= b^2+c^2-a^2 2 bc & (2)/(3)= 8^2+c^2-7^2 2 × 8 × c & C=9 & C=(7^2+8^2-9^2)/(2 × 7 × 8)=(2)/(7) & 49 3 C+42 & 49(4 ^3 C-3 C)+42 & 49(4((2)/(7))^3-3((2)/(7)))+42 & =(32)/(7) & ~m+n=32+7=39 aligned

JEE Main 20246 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 6 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Properties of Triangles

JEE Main 2024 · 6 April · Evening Shift · Mathematics · Properties of Triangles

In a triangle ABC, BC=7, AC=8, AB=α ∈ N and A=(2)/(3) . If 49 (3...

In a triangle

\mathrm{ABC}, \mathrm{BC}=7, \mathrm{AC}=8, \mathrm{AB}=\alpha \in \mathrm{N}

and

\cos \mathrm{A}=\frac{2}{3}

. If

49 \cos (3 \mathrm{C})+42=\frac{\mathrm{m}}{\mathrm{n}}

, where

\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1

, then

\mathrm{m}+\mathrm{n}

is equal to________

Correct answer

Step-by-step solution

\begin{aligned} & \cos \mathrm{A}=\frac{\mathrm{b}^2+\mathrm{c}^2-\mathrm{a}^2}{2 \mathrm{bc}} \\ & \frac{2}{3}=\frac{8^2+\mathrm{c}^2-7^2}{2 \times 8 \times \mathrm{c}} \\ & \mathrm{C}=9 \\ & \cos \mathrm{C}=\frac{7^2+8^2-9^2}{2 \times 7 \times 8}=\frac{2}{7} \\ & 49 \cos 3 \mathrm{C}+42 \\ & 49\left(4 \cos ^3 \mathrm{C}-3 \cos \mathrm{C}\right)+42 \\ & 49\left(4\left(\frac{2}{7}\right)^3-3\left(\frac{2}{7}\right)\right)+42 \\ & =\frac{32}{7} \\ & \mathrm{~m}+\mathrm{n}=32+7=39\end{aligned}

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