In a triangle ABC , with usual notations, if $\mathrm{m} \angle \mathrm{A}=45^{\circ}, \mathrm{m} \angle…

In a triangle ABC , with usual notations, if $\mathrm{m} \angle \mathrm{A}=45^{\circ}, \mathrm{m} \angle \mathrm{B}=75^{\circ}$, then $\mathrm{a}+\mathrm{c} \sqrt{2}$ has the
  1. b
  2. $\frac{\mathrm{b}}{2}$
  3. 2 b
  4. 3 b

Solution

Given that : $\mathrm{m} \angle \mathrm{~A}=45^{\circ}, \mathrm{m} \angle \mathrm{~B}=75^{\circ}, \mathrm{m} \angle \mathrm{C}=60^{\circ}$ we know that $\begin{aligned} & \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c} \\ & \frac{\sin 45^{\circ}}{a}=\frac{\sin 75^{\circ}}{b}=\frac{\sin 60^{\circ}}{c} \\ \therefore \quad & \frac{\sin 45^{\circ}}{a}=\frac{\sin \left(45^{\circ}+30^{\circ}\right)}{b} \end{aligned}$ $\begin{array}{ll}\therefore & \frac{1}{\sqrt{2} a}=\frac{\frac{1}{\sqrt{2}} \times \frac{1}{2}+\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}}{b} \\ \therefore & \frac{1}{\sqrt{2} a}=\frac{1+\sqrt{3}}{2 \sqrt{2} b} \\ \therefore & a=\frac{2 b}{1+\sqrt{3}} \\ \therefore & \frac{\sin 60^{\circ}}{c}=\frac{\sin \left(45^{\circ}+30^{\circ}\right)}{b} \\ \therefore & \sqrt{2} c=\frac{2 \sqrt{3} b}{1+\sqrt{3}} \\ \therefore \quad & a+\sqrt{2} c=\frac{2 b}{1+\sqrt{3}}+\frac{2 \sqrt{3} b}{1+\sqrt{3}}=2 b\end{array}$

Asked in: MHT CET 2024 (10 May Shift 2)

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