Question

In the given figure, ABCD is a cyclic quadrilateral such that AB is a diameter of the circle. If ∠ADC = 130°, find∠CAB.​

Answer 1

Solution: Since ABCD is a cyclic quadrilateral,

∴ ∠ADC +∠CBA =180°

130°+∠CBA =180°

∠CBA =180°-130°

∠CBA =50°

∠ACB=90° [ Angle in a semicircle]

In triangle ACB, we have

∠ACB+ ∠CBA+∠CAB= 180° [Sum of the angles of a triangle is 180°]

90°+50°+∠CAB= 180°

140°+∠CAB= 180°

∠CAB= 180°-140°=40°

Thus,∠CAB=40°

Answer 2

Answer:

40°

Step-by-step explanation:

As ABCD is a cyclic quadrilateral, so ,

ABC + CBA = 180 °

ABC = 130° { GIVEN}

130° + CBA = 180°

CBA = 50°

ACB = 90° { ANGLE IN SEMICIRCLE}

IN ∆ABC we have

ACB + CBA + CAB = 180°

90°+50°+CAB = 180°

140°+CAB = 180°

CAB = 40°