if an angle of a parallelogram is two third of its adjacent angle then find the smallest angle of parallelogram
biluminous:
I am not sure if my solution is wright or wrong.
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2/3x is smallest angle
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in a parallelogram
![x + (2 \div 3)x = 180 x + (2 \div 3)x = 180](https://tex.z-dn.net/?f=x+%2B+%282+%5Cdiv+3%29x+%3D+180)
![(5 \div 3) \times x = 180 (5 \div 3) \times x = 180](https://tex.z-dn.net/?f=%285+%5Cdiv+3%29+%5Ctimes+x+%3D+180)
so,
![x = 108 \: deg x = 108 \: deg](https://tex.z-dn.net/?f=x+%3D+108+%5C%3A+deg)
and
![(2 \div 3)x = (2 \div 3) \times 108 (2 \div 3)x = (2 \div 3) \times 108](https://tex.z-dn.net/?f=%282+%5Cdiv+3%29x+%3D+%282+%5Cdiv+3%29+%5Ctimes+108)
![(2 \div 3)x = 72 \: deg (2 \div 3)x = 72 \: deg](https://tex.z-dn.net/?f=%282+%5Cdiv+3%29x+%3D+72+%5C%3A+deg)
thus smallest angle is 72°.
so,
and
thus smallest angle is 72°.
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