Q25 / 120 · NDA 2024 (Phase 2) · Mathematics

In a triangle $\mathrm{ABC}, \angle \mathrm{A}=75^{\circ}$ and $\angle \mathrm{B}=45^{\circ}$. What is $2 a-b$ equal to?

  • A
    $c$
  • Correct answer
    B
    $\sqrt{2} c$
  • C
    $2 c$
  • D
    $2 \sqrt{2} c$

Solution

Here, $\angle \mathrm{A}=75^{\circ}, \angle \mathrm{B}=45^{\circ} \Rightarrow \angle \mathrm{C}=60^{\circ}$ We know that, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \Rightarrow \frac{a}{\sin 75^{\circ}}=\frac{b}{\sin 45^{\circ}}=\frac{c}{\sin 60^{\circ}}$ Hence, $\begin{aligned} & a=\frac{\sin 75^{\circ}}{\sin 60^{\circ}} c=\frac{\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right) c}{\left(\frac{\sqrt{3}}{2}\right)}=\frac{(\sqrt{3}+1) c}{\sqrt{6}} \\ & \text { and } b=\frac{\sin 45^{\circ}}{\sin 60^{\circ}} \cdot c=\frac{\left(\frac{1}{\sqrt{2}} c\right)}{\left(\frac{\sqrt{3}}{2}\right)}=\frac{2 c}{\sqrt{6}} \end{aligned}$ Now, $2 \mathrm{a}-\mathrm{b}=\frac{2 \sqrt{3}}{\sqrt{6}} \mathrm{c}=\sqrt{2} \mathrm{c}$

Hence, the correct answer is option: 'B'.

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