Question

Two circles with centre P and Q
intersect at B and C. A, D are
points on the circle such that A,
C, D are collinear. If angle APB =
130°, and angle BQD = x, then the
value of xis
(1) 65
(2) 130
(3) 195
(4) 135​

Answer 1

Answer:

$\mathfrak \red{(b) \: 130°}$

Step-by-step explanation:

∠ACB =

$\frac{(360 -∠APB)}{2}$

$\frac{(360 - 130)}{2} = 115$

Now,

$\mathbb{∠BCD = 180 - 115 = 65}$

$\mathbb{∠BQD \: (Inner \: Angle) = 360 - (2 \times 65) = 230}$

$\mathbb{∠BQD \: (Outer \: Angle) = 360 - 230 = 130}$

Answer 2

Answer:

[tex] <marquee > Hope It Helps You❣️✌