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Let A, B, and C be the internal angles of an isosceles triangle. The

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Bunuel
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Let A, B, and C be the internal angles of an isosceles triangle. The [#permalink] New post   08 Jul 2020, 02:22
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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°
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C
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Re: Let A, B, and C be the internal angles of an isosceles triangle. The [#permalink] New post   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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Re: Let A, B, and C be the internal angles of an isosceles triangle. The [#permalink] New post   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
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Re: Let A, B, and C be the internal angles of an isosceles triangle. The [#permalink] New post   08 Jul 2020, 02:44
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Acute angle is always less than 90 degree.
We can eliminate last 2 options.

Take 80 as our max angle.

80/4 = 20 = C.
A = 80
A + B + C =180
A = B as its an iso . triangle

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Re: Let A, B, and C be the internal angles of an isosceles triangle. The [#permalink] New post   08 Jul 2020, 02:46
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Acute angle is always less than 90 degree.
We can eliminate last 2 options.

Take 80 as our max angle.

80/4 = 20 = C.
A = 80
A + B + C =180
A = B as its an isoceles triangle
80 it is.

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Re: Let A, B, and C be the internal angles of an isosceles triangle. The [#permalink] New post   08 Jul 2020, 05:47
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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°

A = x
Case I: C = 4x, B = 4x
x + 4x + 4x = 180°
4x = 80°

Case II: C = x, B = 4x
x + x + 4x = 180°
4x = 120°
This invalidates the conditions

Answer C.
Note: Since triangle is acute, maximum angle value possible is less than 90°. So, D and E are eliminated. As largest angle is to be found, C is largest among option left.
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Re: Let A, B, and C be the internal angles of an isosceles triangle. The [#permalink] New post   08 Jul 2020, 08:05
given all angles are acute i.e <90°
since ∆ ABC is an isoscles angle B = angle C
let angle c = x
so angle a= x/4
we get
x/4+x+x = 180
x=80˚
OPTION C ; degree measure for the largest acute angle of the triangle


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°
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Re: Let A, B, and C be the internal angles of an isosceles triangle. The [#permalink] New post   08 Jul 2020, 08:33
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Two angles of the triangle are equal and angle A is 1/4th of angle C.
Now, angle B can be equal to angle A or C.
Case 1: B is equal to angle A
=> C/4 + C/4 + C = 180 (sum of angles of triangle is 180)
=> 3C/2 = 180
=> C=120
and A=B= 30.
But it is mentioned that all angles are acute hence this case is not possible.
Case 2: B is equal to C
=> C/4 + C+ C=180 (sum of angles of triangle)
=> 9C/4 =180
=> C=80
and B=80 and A=20
Now, question asks largest acute angle which is 80 in this case.
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Re: Let A, B, and C be the internal angles of an isosceles triangle. The [#permalink] New post   08 Jul 2020, 09:03
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Quote:
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°


a,b,c < 90 (acute)
sum = 180
a=1/4c then a=b or b=c
b=c
sum=1/4c+2c=9c/4=180
c=80, b=80, a=20

Ans (C)
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Re: Let A, B, and C be the internal angles of an isosceles triangle. The [#permalink] New post   08 Jul 2020, 09:18
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since triangle is isosceles, let the two angles be x and 4x.
the third angle will be x (since it is given in the question "the" largest acute angle; So, only one angle is largest.
therefore measure of all angles:
180 = x + x + 4x = 6x

solving we get x = 30
therefore, 4x = 120
Answer E.
Notice that if 4x = 80,
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Re: Let A, B, and C be the internal angles of an isosceles triangle. The [#permalink]
08 Jul 2020, 09:18
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