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:
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
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