JEE Main 2015 (Online) 10th AprilQuestion 1
In a $\Delta ABC$, $\frac ab=2+\sqrt3$, and $\angle C=60^\circ$. Then the ordered pair $(\angle A,\angle B)$ is equal to :
Trigonometric Ratio & Identities
Solution
Concept. Napier's analogy links $\frac{a-b}{a+b}$ with $\tan\frac{A-B}{2}$. Formulas used. (1) Napier: $\tan\dfrac{A-B}{2}=\dfrac{a-b}{a+b}\cot\dfrac C2$Step 1 - The ratio. With $\frac ab=2+\sqrt3$: $$\frac{a-b}{a+b}=\frac{1+\sqrt3}{3+\sqrt3}=\frac{1+\sqrt3}{\sqrt3(\sqrt3+1)}=\frac{1}{\sqrt3}.$$ Step 2 - Use (1). $\cot30^\circ=\sqrt3$, so $\tan\frac{A-B}{2}=1$ and $A-B=90^\circ$. Step 3 - Solve. $A+B=180^\circ-60^\circ=120^\circ$, so $A=105^\circ$, $B=15^\circ$. Answer: (A)