Question

find the following. ​

Answer 1

Here, O is actually the circum centre of ∆PAB

i) <APB = 1/2 * <AOB = 1/2 * 126° = 63°

ii) Now, OR ⊥ AB is drawn.

In ∆AOR & ∆BOR,

a) AO = OB (Both are circum radius)

b) <ARO = <ORB (Both are right anglel

c) OR = OR (Common)

=> ∆AOR ≅ ∆BOR (S-A-S)

=> <OAB = <OBA = 29° (C-P-C-T)

=> <AOB = 180°-58°=122°

=> <APB = 1/2 * <AOB = 1/2 * 122° = 61°

iii) <APB = 62°

=> <AOB = 2<APB = 124°

=> <OBA + <OAB = 180°- <AOB = 180° - 124° = 56°

=> 2<OBA = 56° [<OBA = <OAB(From above proof)]

=> <OBA = 28°