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Trapezium Trigonometric Ratio

XAT 2019 · Quantitative Ability & Data Interpretation · Medium · Geometry

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In the trapezium ABCD the sides AB and CD are parallel. The value of sin⁡∠BACsin⁡∠BAD\frac{\sin\angle BAC}{\sin\angle BAD} is

  1. A.

    BC/AD

  2. B.

    AB/CD

  3. C.

    AC/CD

  4. D.

    AD/AC

  5. E.

    AC/AD

Answer

D

Explanation

Let hh be the height of the trapezium. In ΔBAC\Delta BAC, height from C to AB is hh, so sin⁡∠BAC=hAC\sin\angle BAC = \frac{h}{AC}. In ΔBAD\Delta BAD, height from D to AB is hh, so sin⁡∠BAD=hAD\sin\angle BAD = \frac{h}{AD}. Ratio sin⁡∠BACsin⁡∠BAD=h/ACh/AD=ADAC\frac{\sin\angle BAC}{\sin\angle BAD} = \frac{h/AC}{h/AD} = \frac{AD}{AC}.

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