Find out the value of angle A (i.e. angle BAC).?

Question

In a triangle ABC, angle B = 2 x angle C. D is a point on BC such that AB = CD and angle BAD = angle DAC.

Answer 1

The original triangle ABC has angles C = x, B = 2x and A = (180-3x)

Since BAD = DAC, line ad bisects angle A.

Within triangle ADC, you have angles A = 90-3x/2, C = x and D = (90+x/2)

Within triangle ABD, you have angles A = 90-3x/2, B = 2x and D = 90-x/2

In ADC, by the law of sines, (90-3x/2)/cd = x/ad; (90-3x/2)/x = cd/ad

In ABD, by the law of sines, (90-x/2)/ab = 2x/ad; (90-x/2)/2x = ab/ad = cd/ad (recall that ab = cd).

So
(90-3x/2)/x = (90-x/2)/2x
2(90-3x/2) = 90-x/2
2(180-3x) = 180-x
360-6x = 180-x
180 = 5x
36 = x

So
angle C = 36,
angle B = 2*36 = 72,
angle A = 180-(72+36) = 180 - 108 = 72.

Answer 2

See, if the angle C is x then angle B is 2x. then angle A is 180-3x.Angle C is 20 degrees,Angle Bis 40 degrees angle A is 120 degrees.

Answer 3

180 degree

Answer 4

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