In Fig., AD and CF are respectively perpendiculars to sides BC and AB of ∆ABC. If ∠FCD = 50°, find ∠BAD.

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Solution

$We know that the sum of all angles of a triangle is 180 ° . Therefore, for the given ∆ FCB, we can say that : ∠ FCB + ∠ CBF + ∠ BFC = 180 ° \Rightarrow 50 ° + ∠ CBF + 90 ° = 180 ° Or, ∠ CBF = 180 ° - 50 ° - 90 ° = 40 ° . . . (i) Using the above rule for ∆ ABD, we can say that : ∠ ABD + ∠ BDA + ∠ BAD = 180 ° \Rightarrow ∠ BAD = 180 ° - 90 ° - 40 ° = 50 ° [From (i)]$

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