Question

In figure ab=ac,ch=cb and hk parallel bc if angle cax=137 then find angle chk

Answer 1

Solution:-
∠ XAK + ∠ KAH = 180° (LP)
∠ KAH = 180°- 137° = 43°
AB = AC
∠ ABC = ∠ ACB = 137/2 = 68.5°
CH = CB
⇒ ∠ CBA = ∠CHB = 68.5°
∴ ∠ HCB =180°- 137° = 43°
∠ HCB = ∠ CHK = 43° (Alternative angles)
So, angle CHK = 43°

Answer 2

Angle XAC=137

So, Angle B + Angle C = 137 ( SUM of two opposite interior Angle)

And Angle A = Angle B (AB=AC)

Let angle A=B=X So,

X+X=137

2X=137

X=68.5

AngleA= AngleB =68.5

Angle B =Angle CHB (CH=CB) So ,

Angle CHB = 68.5 AND

In triangle HBC

ANGLE HBC= 68.5

ANGLE BHC= 68.5

ANGLE HCB= ? So ,

ANGLE HBC+ANGLE BHC+ANGLE HCB = 180°

68.5+68.5+ANGLE HCB = 180°

137+ANGLE HCB=180°

ANGLE HCB=180+137°

ANGLE HCB=43

ANGLE HCB=43=ANGLE CHK(alternative angle)