Q217 / 270 · NDA 2014 (Phase 2) · Mathematics

In a triangle $\mathrm{ABC}, c=2, \mathrm{~A}=45^{\circ}, a=2 \sqrt{2}$, then what is C equal to?

  • Correct answer
    A
    $30^{\circ}$
  • B
    $15^{\circ}$
  • C
    $45^{\circ}$
  • D
    None of these

Solution

$\begin{aligned} \frac{a}{\sin \mathrm{~A}} & =\frac{b}{\sin \mathrm{~B}}=\frac{c}{\sin \mathrm{C}} \\ \frac{2 \sqrt{2}}{\sin 45^{\circ}} & =\frac{2}{\sin \mathrm{C}}\end{aligned}$ $\begin{array}{ll}\Rightarrow & \frac{2 \sqrt{2}}{\frac{1}{\sqrt{2}}}=\frac{2}{\sin \mathrm{C}} \\ \Rightarrow & 4 \sin \mathrm{C}=2 \\ \Rightarrow & \sin \mathrm{C}=\frac{2}{4}=\frac{1}{2}=\sin 30^{\circ} \\ \Rightarrow & \mathrm{C}=30^{\circ}\end{array}$

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

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