What is \frac{AB}{\sin C} equal to ?

Consider the following for the next three (03) items that follow : ABC is a triangular plot with AB=16\text{ m}, BC = 10\text{ m} and CA=10\text{ m}. A lamp post is situated at the middle point of the side AB. The lamp post subtends an angle 45^{\circ} at the vertex B.

  1. A. 17\text{ m}
  2. B. \frac{50}{3}\text{ m} ✓
  3. C. \frac{40}{3}\text{ m}
  4. D. 16\text{ m}

Correct Answer: B. \frac{50}{3}\text{ m}

Explanation

Let sides be c=16, a=10, b=10. Using Heron's formula, the semi-perimeter s = \frac{10+10+16}{2} = 18. Area \Delta = \sqrt{18(18-10)(18-10)(18-16)} = \sqrt{18 \times 8 \times 8 \times 2} = 48. Also, \Delta = \frac{1}{2}ab\sin C \implies 48 = \frac{1}{2}(10)(10)\sin C \implies \sin C = \frac{48}{50} = \frac{24}{25}. Therefore, \frac{AB}{\sin C} = \frac{16}{24/25} = \frac{16 \times 25}{24} = \frac{50}{3}.

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