Question

In a cyclic quadrilateral ABCD, $\angle A=(2x+4{)}^{\circ },\angle B=(y+3{)}^{\circ },\angle C=(2y+10{)}^{\circ },\angle D=(4x-5{)}^{\circ }$. Find the four angles.

Answer 1

We know that the sum of the opposite angles of cyclic quadrilateral is 180^o. In the cyclic quadrilateral ABCD, angles A and C and angles B and D are pairs of opposite angles.

Therefore ∠A + ∠C = 180^o and ∠B + ∠D = 180^o

Taking ∠A + ∠C = 180^o

By substituting ∠A = (2x + 4)^o and ∠C = (2y + 10)^o we get

2x + 4 + 2y + 10 = 180^o

2x + 2y + 14 = 180^o

2x + 2y = 180^o – 14^o

2x + 2y = 166 —— (i)

Taking ∠B + ∠D = 180^o

By substituting ∠B = (y+3)^o and ∠D = (4x – 5)^o we get

y + 3 + 4x – 5 = 180^o

4x + y – 5 + 3 = 180^o

4x + y – 2 = 180^o

4x + y = 180^o + 2^o

4x + y = 182^o ——- (ii)

By multiplying equation (ii) by 2 we get 8x + 2y = 364 —— (iii)

By subtracting equation (iii) from (i) we get

-6x = -198

x=−198−6x=−198−6

x = 33^o

By substituting x = 33^o in equation (ii) we get

4x + y = 182

132 + y = 182

y = 182 – 132

y = 50^o

The angles of a cyclic quadrilateral are

∠A = 2x + 4

= 66 + 4

= 70^o

∠B = y + 3

= 50 + 3

= 53^o

∠C = 2y + 10

= 100 + 10

= 110^o

∠D = 4x – 5

= 132 – 5

= 127^o

Hence, the angles of cyclic quadrilateral ABCD are ∠A = 70^o, ∠B = 53^o, ∠C = 110^o, ∠D = 127^o