P is a point on line AB and PQ is a ray such that ∠QPA =7x and ∠QPB = 5x. Then, the value of (8x-10°) is:

Correct option is C

P is a point on line AB
PQ is a ray
∠QPA = 7x
∠QPB = 5x
We need to find the value of (8x-10°)
Since P is on line AB, the sum of the angles at point P is 180°. Therefore, we can set up an equation:
∠QPA + ∠QPB + ∠P = 180°
Substituting the given values:
7x + 5x + ∠P = 180°
12x + ∠P = 180°
12x = 180° - ∠P
Now, consider triangle PQA:
∠QPA = 7x
∠A = 180° - ∠QPA - ∠P
∠A = 180° - 7x - ∠P
The sum of the angles in a triangle is 180°, so we can set up an equation for triangle PQA:
∠QPA + ∠A + ∠P = 180°
7x + 180° - 7x - ∠P + ∠P = 180°
180° - ∠P = 180°
-∠P = 0
Therefore, ∠P = 0°
Substituting this value back into the equation 12x = 180° - ∠P:
12x = 180° - 0°
12x = 180°
x = 15°
Now we can calculate the value of (8x-10°):
(8x-10°) = (8 × 15°) - 10°
(8x-10°) = 120° - 10°
(8x-10°) = 110°
Therefore, the value of (8x-10°) is 110°. So the answer is (c) 110°.