MHTCET 19 Apr 2026 Shift 2 Paper

Concept:Use Heron's formula to rewrite the given product as $16\Delta^2$, then compare it with the area formula $\Delta = \frac{1}{2}bc\sin A$.Explanation:For $\triangle ABC$, Heron's formula gives the area as$$\Delta = \frac{1}{4}\sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}$$Squaring both sides, we obtain$$16\Delta^2 = (a+b+c)(b+c-a)(c+a-b)(a+b-c)$$The given condition states that$$(a+b+c)(b+c-a)(c+a-b)(a+b-c) = 3b^2c^2$$Hence,$$16\Delta^2 = 3b^2c^2$$The area using sides $b$, $c$ and the included angle $A$ is$$\Delta = \frac{1}{2}bc\sin A$$Substituting this into the previous equation,$$16\left(\frac{1}{2}bc\sin A\right)^2 = 3b^2c^2$$Simplifying,$$4b^2c^2\sin^2 A = 3b^2c^2$$Since $b^2c^2 \neq 0$, cancel it to get$$\sin^2 A = \frac{3}{4}$$For a triangle, $0^\circ < A < 180^\circ$, so $\sin A$ is positive:$$\sin A = \frac{\sqrt{3}}{2}$$Therefore,$$A = 60^\circ \quad \text{or} \quad 120^\circ$$Answer:Option A: $60^\circ$ or $120^\circ$