Find question attached.
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a.x=50°
b.x=22°
Answer:
Solution given:
a:
3x=2*75°[inscribed angle is half of central angle]
3x=150°
x=150°/3=50°
x=50°
b.
<BDC=34°+x[exterior angle is equal to the sum of two opposite interior angle of triangle]
again
<DCB=34°+x[base angle of isosceles triangle]
again
<ABC=90°[inscribed angle on a semi circle is 90°]
Now.
In triangle
ABC
<A+ <B+<C=180°[sum of interior angle of a triangle is 180°]
34°+90°+34°+x=180°
x=180°-90°-68°
x=90°-68°
x=22°
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.