Question

6 .If AB = AC.CH = CB and HK II BC .And <CAX =137° then find< CHK
A
137
H
K к
BE
<
CHK ??
​

Answer 1

Answer:

CHK=43°

Step-by-step explanation:

<XAK+<KAH=180°(LP)

<KAH=180°-137°=43°

AB=AC

<ABC=<ACB=137/2=68.5°

CH=CB

=> <CBA =<CHB =68.5°

i.e <HCB =180°-137°=43°

<HCB =<CHK = 43°(Aternative angles)

So, angle CHK = 43°