In the figure given below AB = AC, CH = CB and HK ║BC.If the exterior angle CAX is 140^o ,then

Question

In the figure given below AB = AC, CH = CB and HK
â•‘BC If the exterior angle CAX is
140o ,then

Ajanta Trivedi · Accepted Answer

given: AB=AC and CH=CB, and HK is parallel to BC,

exterior angle CAX=140

to find the angle HCK

∠BAC=180-∠CAX [angles on the same side of the straight line are supplementary]

∠BAC=180-140=40

since AB=AC

in the triangle ABC, ∠ABC=∠ACB (angles opposite to equal sides are equal in length)

therefore ∠ABC=∠ACB= $\frac{1}{2} (180 - 40) = \frac{1}{2} * 140 = 70$ .........(1)

now in the triangle HBC, CH=CB⇒∠HBC=∠BHC

∠HBC=∠ABC=70=∠BHC..........(2)

now in the triangle HCB,

∠HCB=180-(70+70)=180-140=40

∠HCK=∠ACB-∠HCB=70-40=30

In the figure given below AB = AC, CH = CB and HK ║BC.If the exterior angle CAX is 140o ,then