Answer:
Explanation:
You want the obtuse angle BCH in a figure with parallel lines GE and HF crossed by transversal BC, where the obtuse exterior angle at B is marked 10x+5, and the acute exterior angle at C is marked 6x+5.
a) Consecutive exterior angles
The two marked angles are "consecutive exterior angles". As such, they are supplementary:
(10x +5) +(6x +5) = 180
16x = 170 . . . . . . . . . . . . . . subtract 10
x = 170/16 = 10 5/8 = 10.625
b) Obtuse angle
All of the obtuse angles in the figure have same measure, so angle BCH is ...
∠BCH = 10(10.625) +5 = 111.25 . . . . degrees
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