Question

In the given figure, if C is the centre of the circle and PQC = 25° and PRC = 15°, then QCR is equal to :

(A) 40°
(B) 60°
(C) 80°
(D) 120​

Answer 1

(1)

In triangle PQC,

PC = QC.

∠CQP = ∠CPQ

∠CPQ = 25

(2)

In triangle PRC,

PC = RC

∠CRP = ∠CPR

∠CPR = 15

Now,

∠QPR = ∠QPC + ∠CPR

          = 25 + 15

          = 40

Thus,

∠QCR = 2 * ∠QPR

          = 2 * 40

          = 80°

Hope this helps!

Answer 2

Answer:

80 degrees

Step-by-step explanation:

Triangle PQC and triangle PCR are isosceles triangles, that means two sides are equals .

In triangle PQC, sides PC and CQ are equal (Since they are radii of the circle with center C )

Therefore, the angles they subtend are equal, this implies that angle CPQ = 25 degrees

Since the sum of the angles in a triangle = 180 degree, angle PCQ = 180-(25+25) = 130 degrees

Similarly, in triangle PCR,angle RPC= 15 degrees

Therefore angle PCR = 180 - (15+15) = 150 degrees

Total angle around point C = Angle QCR + Angle PCR + Angle PCQ = 360 degrees

Angle QCR = 360 - (150+130) = 80