Question

ABCDisa cyclic quadrilateral.​

Answer 1

Step-by-step explanation:

(a)

Using cosine rule:

$using \: cosine \: rule \\ {ac}^{2} = {ab}^{2} + {bc}^{2} - 2 .ab.bc .\cos( < abc) \\ {ac}^{2} = {9.5}^{2} + {11.1}^{2} - 2 \times 9.5 \times 11.1 \times \cos( 70) \\ {ac}^{2} = 141.32\\ ac = \sqrt{141.32} \\ ac= 11.88$

(b)

(c)

from question (b) angle ADC is 110 degree.

then,

ACD=180-110-37

=33

question (a) ac=11.88

using sine rule:

$\frac{ac}{ \sin(adc)} = \frac{ad}{ \sin(acd)} \\ \frac{11.88}{ \sin(110)} = \frac{ad}{ \sin(33)} \\ ad = \frac{11.88 \times \sin(33)}{ \sin(110)} \\ ad = 6.89$

(d)

(i)

angle ABC and angle AEC are on the same chord of corcle ABECD.

that's why angle ABC is equal to angle AEC.

AEC=70 degree.

(ii)