Question

In the given figure, ABCDE is a pentagon inscribed in a circle. If AB=BC=CD,angle BCD=110° and angleBAE=120°, find
1)angleABC,
2)angleCDE,
3)angle AED,
4)angleEAD

Answer 1

according to the figure,

if we join BD.

In triangle BCD,

BC=BD

____________________________________________________________

<CBD = <CDB = x Isosceles triangle

<BCD=110°,

<BCD+ <CBD + <CDB = 180,

x+x+110=180

2x=70

x=35

Hence,

arc BC = arc CD = arc AB = 2<CBD = 70°

angle BAE=120°

arc BCDE= 2x120° = 240°

arc BC + arc CD + arc DE = 240

70+70+ arc DE = 240

arc DE = 100

<BCD = 2 (Arc DE + Arc EA +Arc AB)

110*2 =2(100+70+Arc EA)

Arc EA=50

<ABC = 1/2 * (arc CD + arc DE arc EA)

<ABC = 1/2* (70+100+50)

<ABC = 110

and similarly you can find for remaining angles.....

Answer 2

Answer:

108

Step-by-step explanation:

(2n-4)x90=(2x5-4) 90=540

x = 540/5=lo8 degree

OR

External angle sum=360

360/5=72 degree

internal angle x=l80-72=108 degree