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Ratio of Altitudes in Triangle

CAT 2022 Slot 2 · Quantitative Ability · Medium · Geometry

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In triangle ABCABC, altitudes ADAD and BEBE are drawn to the corresponding bases. If ∠BAC=45∘\angle BAC = 45^\circ and ∠ABC=θ\angle ABC = \theta, then ADBE\frac{AD}{BE} equals

  1. A.

    2sin⁡θ\sqrt{2}\sin\theta

  2. B.

    2cos⁡θ\sqrt{2}\cos\theta

  3. C.

    sin⁡θ+cos⁡θ2\frac{\sin\theta + \cos\theta}{\sqrt{2}}

  4. D.

    1

Answer

A

Explanation

In △ABC\triangle ABC, area of △ABC=12×BC×AD=12×AC×BE\triangle ABC = \frac{1}{2} \times BC \times AD = \frac{1}{2} \times AC \times BE. Therefore, ADBE=ACBC\frac{AD}{BE} = \frac{AC}{BC}. By Sine Rule in △ABC\triangle ABC: ACsin⁡θ=BCsin⁡45∘  ⟹  ACBC=sin⁡θsin⁡45∘=2sin⁡θ\frac{AC}{\sin\theta} = \frac{BC}{\sin 45^\circ} \implies \frac{AC}{BC} = \frac{\sin\theta}{\sin 45^\circ} = \sqrt{2}\sin\theta. Hence, ADBE=2sin⁡θ\frac{AD}{BE} = \sqrt{2}\sin\theta.

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