a) Solution
By using the inscribed angle is half of central angle,
→ 3x = 2 × 75
→ 3x = 150
→ x = 150/3
→ x = 50°
Thus, 50° is the value of x.
b) Solution
By using the exterior angle is equal to sum of two opposite interior angle of triangle,
→ <BDC = 34+x
→ <DCB = 34+x
(base angle of isosceles triangle)
→ <ABC = 90°
(inscribed angle on a semi circle is 90°)
Then in ∆ ABC,
By sum of interior angle of a triangle is 180°,
→ <A+<B+<C = 180°
→ 34+90+34+x=180°
→ x = 180°-90°-68°
→ x = 90°-68°
→ x = 22°
Thus, 22° is the value of x.