Answer:
(a) ∠DAB = 59°
(b) ∠A = 135°
(c) ∠A = 99°
Step-by-step explanation:
Theorem: If a quadrilateral is inscribed inside a circle, then the opposite angles of the quadrilateral are SUPPLEMENTARY.
So, here in the given figures:
(1) ∠DCB + ∠DAB = 180° (as opposite angles are supplementary)
⇒∠DAB = 180° - 121° = 59°
or, ∠DAB = 59°
(2) ∠DCB + ∠DAB = 180° (as opposite angles are supplementary)
⇒(3x+6) + (x+2) = 180° or, 4x = 180 - 8 = 172
⇒ x = 172/4 = 43, or x = 43
So, ∠A = (3x+6) = 3(43) + 6 = 135°
(3) (28) + (x) = 180°
or, x = 180 - 28 = 152
So, ∠A = (x- 36) = 135 - 36 = 99°
(4)
Answer:
Step-by-step explanation:
1. Opposite angles are supplementary
A+C = 180
A = 180-121 = 59
2. Opposite angles are supplementary
A+C = 180
3x+6+x+2 = 180
4x+8 = 180
4x = 172
x = 43
A = 3x+6 = 3(43)+6
= 135
3. Opposite angles are supplementary
B+D = 180
28+x = 180
x = 152
A = x-36 = 152-36
= 116
4. Opposite angles are supplementary
f+90 = 180
f = 90
Central angle = 2x inscribed angle = 2x75 = 150
Sum of central angles = 360
150+110+g = 360
g = 360-150-110 = 100
5. Opposite angles are supplementary
S+V = 180
x+77+x+103=180
2x+180=180
x=0
S=77
Central angle = 2x inscribed angle = 2x77 = 154
Sum of central angles = 360
mRST+154 = 360
mRST = 360-154 = 206
