JEE Advanced 2009 Paper 2Question 1
Let $ABC$ and $ABC'$ be two non-congruent triangles with sides $AB=4$, $AC=AC'=2\sqrt2$ and angle $B=30^\circ$. The absolute value of the difference between the areas of these triangles is
Trigonometric Ratio & Identities
Answer$4$
Solution
Concept. Two sides and a non-included angle can give two triangles: the cosine rule at $B$ is a quadratic in $BC$ with two positive roots. Formulas used. (1) $AC^2=AB^2+BC^2-2\cdot AB\cdot BC\cos B$(2) Area $=\frac12\cdot AB\cdot BC\sin B$Step 1 - The two triangles.Step 2 - Cosine rule. With $BC=a$, by (1): $$8=16+a^2-4\sqrt3\,a\ \Longrightarrow\ a^2-4\sqrt3\,a+8=0\ \Longrightarrow\ a=2\sqrt3\pm2.$$ Step 3 - Areas. By (2), area $=\frac12\cdot4\cdot a\cdot\frac12=a$. So the areas are $2\sqrt3+2$ and $2\sqrt3-2$. Step 4 - Difference. $(2\sqrt3+2)-(2\sqrt3-2)=4$. Answer: 4