Discussion

Question:

In triangle ABC, altitudes AD and BE are drawn to the corresponding bases. If ∠BAC = 45° and ∠ABC = θ, then AD/BE equals

Explanation:

Area of ∆ABC = $\frac{1}{2}$ × AB × AC × sin 45° = $\frac{A B \times A C}{2 \sqrt{2}}$ ...(1)

Area of ∆ABC = $\frac{1}{2}$ × BA × BC × sin θ° = $\frac{B A \times B C \times \sin θ}{2}$ ...(2)

(1) = (2)

⇒ $\frac{A B \times A C}{2 \sqrt{2}}$ = $\frac{B A \times B C \times \sin θ}{2}$

⇒ $\frac{A C}{B C}$ = √2 × sin⁡𝜃 ...(3)

Area of ∆ABC = $\frac{1}{2}$ × AD × BC = $\frac{1}{2}$ × AC × BE

⇒ $\frac{A D}{B E}$ = $\frac{A C}{B C}$

⇒ $\frac{A D}{B E}$ = √2 × sin⁡𝜃

Hence, option (c).

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