Q191 / 270 · NDA 2022 (Phase 2) · Mathematics

Paragraph: $A B C$ is a triangular plot with $A B=16 \mathrm{~m}, B C=10 \mathrm{~m}$ and $C A=10 \mathrm{~m}$. A lamp post is situated at the middle point of the side $A B$. The lamp post subtends an angle $45^{\circ}$ at the vertex $B$.
Question: What is $\frac{A B}{\sin C}$ equal to?

  • A
    17 m
  • Correct answer
    B
    $\frac{50}{3} \mathrm{~m}$
  • C
    $\frac{40}{3} \mathrm{~m}$
  • D
    16 m

Solution

$\begin{aligned} & \cos C=\frac{a^2+b^2-c^2}{2 a b}=\frac{100+100-256}{200}=\frac{-56}{200}=\frac{-7}{25} \\ & \sin C=\sqrt{1-\cos ^2 C}=\sqrt{1-\frac{49}{625}}=\frac{24}{25} \\ & \therefore \frac{A B}{\sin C}=\frac{16 \times 25}{24}=\frac{50}{3} \mathrm{~m}\end{aligned}$

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

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