Question

O is the centre of a circle and AB is a diameter, ABCD is a cyclic Quadrilateral, angle ABC=65 degree, angle DAC=40 degree, then measure of angle BCD is​

Answer 1

Answer : 115° Explaination:

Given:

∠ABC=65°∠DAC=40°∠BCD= ??∵

∠ACB is angle in a semicircle.∴

∠ACB = 90°∵ ∠ABC and ∠ADC are supplementary angles ( as the sum of opposite angles of a cyclic quadrilateral is 180° ).∴ ∠ABC + ∠ADC = 180°⇒ 65°+∠ADC = 180°⇒ ∠ADC = 180°- 65°⇒ ∠ADC = 115°In Δ ADC,∠ADC = 115°, ∠DAC = 40°

∴ ∠DAC = 180°-∠ADC-∠DAC= 180°- 115°- 40°= 65°- 40°= 25°

∴ ∠BCD = ∠ACB+∠DCA= 90°+25°= 115°

Answer 2

Answer:

115°

Step-by-step explanation:

Given:

∠ABC=65°

∠DAC=40°

∠BCD= ??

∵ ∠ACB is angle in a semicircle.

∴ ∠ACB = 90°

∵ ∠ABC and ∠ADC are supplementary angles ( as the sum of opposite angles of a cyclic quadrilateral is 180° ).

∴ ∠ABC + ∠ADC = 180°

⇒ 65°+∠ADC = 180°

⇒ ∠ADC = 180°- 65°

⇒ ∠ADC = 115°

In Δ ADC,

∠ADC = 115°, ∠DAC = 40°

∴ ∠DAC = 180°-∠ADC-∠DAC

= 180°- 115°- 40°

= 65°- 40°

= 25°

∴ ∠BCD = ∠ACB+∠DCA

= 90°+25°

= 115°