3ab928654.webp" data-orsrc="https://uploads-pracbee-ramanuj.s3.ap-south-1.amazonaws.com/pyq-question-images/batch-001/8293ab928654.webp" loading="lazy" style="max-width:100%;height:auto;display:block;margin:0 auto;max-height:40vh;vertical-align:baseline" alt="JEE Main 2024 (Online) 4th April Evening Shift Mathematics - Properties of Triangle Question 4 English Explanation 1">

$$\begin{aligned} & P=\frac{|2-1-4|}{\sqrt{1^2+1^2}}=\frac{3}{\sqrt{2}} \\ & \sin \left(\frac{\pi}{6}\right)=\frac{3 / \sqrt{2}}{A B}=\frac{1}{2} \Rightarrow A B=\frac{6}{\sqrt{2}} \\ & \Rightarrow(\alpha-1)^2+(\alpha+4-2)^2=18 \end{aligned}$$

$$\Rightarrow 2 \alpha^2+2 \alpha-13=0 \rightarrow \alpha$$ and $$\gamma$$ satisfy same equation

JEE Main 2024 (Online) 4th April Evening Shift Mathematics - Properties of Triangle Question 4 English Explanation 2

$$\begin{aligned} & \Rightarrow \alpha^2+\gamma^2=(\alpha+\gamma)^2-2 \alpha \gamma \\ & =(-1)^2-2\left(\frac{-13}{2}\right)=1+13=14 \end{aligned}$$

19:T621,

JEE Main 2024 (Online) 4th April Evening Shift Mathematics - Properties of Triangle Question 4 Hindi Explanation 1

$$\begin{aligned} & P=\frac{|2-1-4|}{\sqrt{1^2+1^2}}=\frac{3}{\sqrt{2}} \\ & \sin \left(\frac{\pi}{6}\right)=\frac{3 / \sqrt{2}}{A B}=\frac{

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