In an isosceles triangle, if the vertex angle is twice the sum of the base angles, then the angles of the triangle are

90^o, 45^o, 45^o

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120^o, 30^o, 30^o

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60^o, 60^o, 60^o

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80^o, 50^o, 50^o

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Solution

The correct option is B
120^o, 30^o, 30^o

Let the base angles be x and x.

Then, vertical angle = 4x.

Now, x + x + 4x = 180° (angle sum property of a triangle)

therefore x = 30°

Hence, angles of the triangle are 120°, 30°, 30°.

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Similar questions

a) {90^{o}, 30^{o}, 50^{o}}

b) {40^{o}, 40^{o}, 100^{o}}

c) {50^{o}, 60^{o}, 70^{o}}

d) {20^{o}, 30^{o}, 40^{o}}

\frac{{tan}^{2} 60^{o} + 4 {sin}^{2} 45^{o} + {sin}^{2} 90^{o}}{3 {sec}^{2} 30^{o} + {csc}^{2} 60^{o} - {cot}^{2} 30^{o}}

\frac{{tan}^{2} 60^{o} + 4 {cos}^{2} 45^{o} + 3 {sec}^{2} 30^{o} + 5 {cos}^{2} 90^{o}}{c o sec 30^{o} + sec 60^{o} - {cot}^{2} 30^{o}}

\frac{x {cosec}^{2} 30^{o} {sec}^{2} 45^{o}}{8 {cos}^{2} 45^{o} {sin}^{2} 90^{o}} = {tan}^{2} 60^{o} - {tan}^{2} 45^{o}

x

30^{o}, 60^{o}

90^{o}

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