The measures of the exterior angles of an octagon are x°x°, 2x°2x°, 4x°4x°, 5x°5x°, 6x°6x°, 8x°8x°, 9x°9x°, and 10x°10x°. Solve for x.
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The sum of the exterior angles of any polygon is always equal to 360.
Exterior angles are (6x−1)
∘
,(10x+2)
∘
,(8x+2)
∘
,(9x−3)
∘
,(5x+4)
∘
and(12x+6)
∘
Now,
(6x−1)
∘
+(10x+2)
∘
+(8x+2)
∘
+(9x−3)
∘
+(5x+4)
∘
+(12x+6)
∘
=360
o
=>(50x+10)
o
=360
o
=>x=7
Each Exterior angle
=>(6x−1)
o
=6×7−1=41
o
=>(10x+2)
o
=10×7+2=72
o
=>(8x+2)
o
=8×7+2=58
o
=>(9x−3)
o
=9×7−3=60
o
=>(5x+4)
o
=5×7+4=39
o
=>(12x+6)
o
=12×7+6=90
o
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