TS ICET 23 May 2019 Shift 2 Paper

Question : 136 of 200

Marks: +1, -0

In a ΔABC , ∠B = 55°, ∠C = 45° and bisector of ∠A meets BC at a point D. Then ∠CAD =

Solution:

In ΔABC

∠ A =180^{∘}-(∠ B +∠ C )

=80^{∘}

∵ AD is the bisector of ∠A

∠ CAD =∠ DAB =\;{80}/{2}=40^{∘}

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