Question

Find the four angles of a cyclic quadrilateral \( ABCD \) in which \( \angle A =(2 x-10)^{\circ}, \angle B =(2 y-20)^{\circ} \), \( \angle C =(2 y+30)^{\circ}, \angle D =(3 x+10)^{\circ} \).

Answer 1

Given,

∠A = (2x−10)°,

∠B = (2y−20)°,

∠C = (2y+30)° and 

∠D = (3x+10)°.

We know, the sum of the opposite angles of a cyclic quadrilateral is 180°.

So ∠A + ∠C = 180°

⇒ (2x−10+2y+30)° = 180°

⇒ (2x+2y)° = 160° −−−(1)

& also, ∠B + ∠D = 180°

⇒ (2y−20+3x+10)° = 180°

⇒ (3x+2y)° = 190∘ −−−−(2)

Subtracting (1) from (2), we  get,

(3x−2x)° = 190° − 160°

⇒ x° = 30°.

Using x° = 30° in (1), we  get,

(2y+2×30)° = 160°

⇒ (2y+60)° = 160°

⇒ (2y)° = 100°

⇒ y° = 50°.

So, ∠A = (2x−10)° = (2×30−10)° = (60−10)° = 50°

∠B = (2y−20)° = (2×50−20)° = (100−20)° = 80°

∠C = (2y+30)° = (2×50+30)° = (100+30)° = 130° 

and ∠D = (3x+10)° = (3×30+10)° = (90+10)° = 100°.

Hence, the smallest angle is ∠A = 50°.