CTET 2 Math and Science Feb 2015 Paper

Question : 38 of 120

Marks: +1, -0

In a quadrilateral ABCD, ∠ D = 60° and ∠ C = 100°. The bisectors of ∠ A and ∠ B meet at the point P. The measure of ∠ APB is

Solution:

We know that, the sum of quadrilateral 360°

∠A+∠B+∠C+∠D=360^{∘}

⇒∠A+∠B+160^{∘}=360^{∘}

⇒∠A+∠B=360^{∘}-160^{∘}

⇒∠A+∠B=200

Now, according to question

∠{PAB}+∠{PBA}+∠{APB}=180^{∘}

⇒\;\;{∠DAB}/{2}+\;\;{∠CAB}/{2}±∠APB=180^{∘}

[∴∠{DAB}=\;\;{∠{A}}/{2}\;\text"and"\;∠{CBA}=\;\;{∠{B}}/{2}]

⇒\;\;{∠{A}+∠{B}}/{2}+∠{APB}=180^{∘}

⇒200+2∠APB=360^{∘}

∠{APB}=\;\;{160^{∘}}/{2}=80^{∘}

Go to Question:

Prev Question

Next Question