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08 Jul 2020, 02:22
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C
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Let A, B, and C be the internal angles of an isosceles triangle. The degree measure of A is one-fourth the degree measure of C. If all the interior angles are acute, what is the degree measure for the largest acute angle of the triangle?
A. 20°
B. 30°
C. 80°
D. 90°
E. 120°
yashikaaggarwal
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08 Jul 2020, 02:30
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Sum of interior angle of triangle = 180
Three angles = X,Y,Z where one of them is equal to other because of isosceles triangle and third angle is 1/4 of other two.
Therefore,
X+X+(X/4) = 180
(9X+X)/4 = 180
X = 80
Answer is C
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Three angles = X,Y,Z where one of them is equal to other because of isosceles triangle and third angle is 1/4 of other two.
Therefore,
X+X+(X/4) = 180
(9X+X)/4 = 180
X = 80
Answer is C
Posted from my mobile device
08 Jul 2020, 02:35
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Explanation:
Acute angles measure less than 90 degrees.
as angle A = C/4.
Either A=B or B=C as ABC is isosceles triangle
We check both cases.
1. A=B=C/4,
So C/4+ C/4 + C = 180
3C/2 =180
C=120 degree which cannot be possible as all angle are below 90 degree.
2. B=C,
C/4+ C + C = 180
C = 80, yes as less than 90 degree
IMO-C
Acute angles measure less than 90 degrees.
as angle A = C/4.
Either A=B or B=C as ABC is isosceles triangle
We check both cases.
1. A=B=C/4,
So C/4+ C/4 + C = 180
3C/2 =180
C=120 degree which cannot be possible as all angle are below 90 degree.
2. B=C,
C/4+ C + C = 180
C = 80, yes as less than 90 degree
IMO-C
