Re: If in the triangle shown above all three interior angles are multiples
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08 Sep 2023, 08:42
Alternative to the algebra, you can plug in numbers. You know that all interior angles are multiples of 10, so x needs to be a multiple of 10. There's an upper limit of 3x to be less than 180, so the max that x can equal is 60 (180/3=60).
This leaves only a couple options for x: x = 10, 20, 30, 40 or 50.
If x = 10:
angle ACD = 30, ACB = 150 (not good, because this would make it the largest angle)
If x = 20:
angle ACD = 60, ACB = 120, CAB = 40, CBA = 30 (not good, ACB is the largest angle)
If x = 30:
ACD = 90, ACB = 90, CAB = 60, CBA = 30 (not good)
x = 40:
ACD = 120, ACB = 60, CAB = 80, CBA = 40 (good)
x = 50:
ACD = 150, ACB = 30, CAB = 100, CBA = 50 (not good, ACB is the smallest)
This means ACB is only the middle angle as a multiple of 10 one time. Choice B is correct.