Question

An exterior angle of a triangle is 130 degrees , then the angle between the bisectors of the other two angles can be

Answer 1

Answer:

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GIVEN: Angle A = 130°

TO FIND : Angle DPE = Angle CPB =?

Let angle B be 2x

So, angle C = 180 - (130 +2x) = 50–2x

So angle DCB =(50–2X)/2

Now, angle CDA = 2x + (50–2x)/2 ( as exterior angle = the sum of 2 opposite interior angles)

=> angle CDA = (2x +50)/2 = x+ 25 ……..(1)

Now,in triangle AEB, angle AEB =180 - (130+x) = 50 -x ………….(2)

So, in quadrilateral ADPE,

angle DPE = 360 -(130 + x+25 + 50-x)

=> angle DPE = 360- 205

=> angle DPE = 155°

Answer 2

Step-by-step explanation:

Solution :

Let three angles of the triangle are A, B and C

Let C= 130 (One angel is given as 130)

Then A+B = 180-130 = 50 (since A+B+C = 180)

Now

C + A/2 + B/2 = 180 (A/2 and B/2 are the bisector)

=> C + (A+B)/2 = 180

=> C + 50/2 = 180

=> C + 25 = 180

=> C = 180-25

=> C = 155