From a point X on a bridge across a river, the angles of depression of two points P and Q on the banks on opposite side of the river are \alpha and \beta respectively. If the point X is at a height h above the surface of the river, what is the width of the river if \alpha and \beta are complementary?

  1. A. 2h(\tan \alpha+\cot \alpha)
  2. B. h \tan \alpha \cdot \tan \beta
  3. C. h \cot \alpha \cdot \cot \beta
  4. D. h \sec \alpha \cdot \csc \alpha

Correct Answer: D. h \sec \alpha \cdot \csc \alpha

Explanation

The width of the river is h \cot \alpha + h \cot \beta. Since \alpha + \beta = 90^\circ, \cot \beta = \tan \alpha. Thus, the width is h(\cot \alpha + \tan \alpha) = h(\frac{\cos \alpha}{\sin \alpha} + \frac{\sin \alpha}{\cos \alpha}) = \frac{h}{\sin \alpha \cos \alpha} = h \sec \alpha \csc \alpha.

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