Question

In the figure, AB is diameter of a circle with centre O and QC is a tangent to the circle at C. If angle CAB= 30degree, find 1. angle CQA 2. angle CBA.

Answer 1

<CAB=30
<ACB=90( ANGLE IN A SEMI CIRLCE)
THEREFORE
<ABC=180-(90+30)
= 180-120
=60

Answer 2

Answer:

CQA = 60° and CBA = 60°

Step-by-step explanation:

Angle ACB = 90 ° (semi circle angle)

therefore,

angle CBA = 180 - (90+30) = 60°

QC is parallel to AB and AC intersects it.

therefore, angle CAB = angle ACQ = 30°

angle QAC = angle ACB = 90°

therefore angle CQA = 180 - (90+30) = 60°