Question

in the figure given below AB=AC ,Ch=CB and HK is parallel to BC if angle C A X is 140 degree then angle H C K is?
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Answer 1

Answer:

Step-by-step explanation:

In the adjoining figure

AB=AC,CH=CB

HK∥BC

If ∠CAD=137

∘

, find ∠CHK

Given : AB=AC,CH=CB and HK∥BC

TO FIND: ∠CHK

SINCE AB=AC

THUS, ΔABC IS ISOSCELES.

∠DAC IS THE EXTERNAL ANGLE, WHICH EQUALS SUM OF OPPOSITE INTERIOR ANGLES:

∠DAC=∠ABC+∠ACB

⇒137

0

=2∠ABC

[∠ABC=∠ACB]

⇒∠ABC=68.5

0

SINCE HK∥BC,∠KHB+∠HBC=180

0

WHICH IMPLIES ∠KHB=180

0

−68.5

0

=111.5

0

SINCE CH=CB, THUS ΔCHB IS ISOSCLELES AND THUS,

∠CHB=∠ABC=68.5

0

SINCE THE ANGLES ARE ADJACENT, ∠CHB+∠CHK=∠BHK

∠CHK=111.5

0

−68.5

0

=43

0

∠CHK=43

0