Question

In the figure given alongside ABCD is a kite whose diagonals AC and BD intersect each other at O. If angle ABO = 32° and angle OCD = 40°, find (i)angle ABC (ii) angle ADC (iii) angle BAD (iv) angle BCD.​

Answer 1

Answer:

ABC = 64°

ADC = 100°

BAD = 98°

BCD = 98°

Step-by-step explanation:

ΔAOB = ΔCOB = 32°

Then,

ABC = 64°

ΔAOB+ΔCOB = 32°+32° = 64° = ABC

ΔADC = 100°

angle C = angle A = 40°

angle D = 180°- ( angle C + angle A )

= 180°-( 40° + 40° )

= 180°- 80°

= 100°

ΔBAD = 98°

angle B = 32°

angle O = 90°

ΔBAO = 180°-(32° + 90°)

= 180°-122°

= 58°

Then angle A = 40° + 58°

= 98°

ΔBCD = ΔBAD = 98°

Answer 2

Answer:

(i) ABC = 64°

(ii) ADC = 100°

(iii) BAD = 98°