If isosceles triangle ABC has a 130° angle at vertex B, which statement must be true? O mZA= 15° and mzC= 35
O mZA+ mZB = 155°
O mZA+ mzC = 60
OMZA= 20° and m2C=30

The true statement about the given angles of the triangle is m∠A + m∠B = 155⁰.

The sum of angles of Isosceles triangle must add up to 180 degrees.

Let the base angles = x and x

x + x + 130 = 180 (sum of angles in a triangle)

2x + 130 = 180

2x = 180 - 130

2x = 50

x = 25

Given angle B = 130°

B + A = B + C = 130 + 25 = 155°

Thus, the true statement about the given angles of the triangle is m∠A + m∠B = 155⁰.

Learn more about angles of Isosceles triangle here: link

Option B m<ZA+m<ZB = 155° is correct option.

Step-by-step explanation:

We know that sum of angles of triangle is equal to 180°

And Isosceles triangle, 2 angles are same

So, let  Angle 1 = x

Angle 2 = x

Angle 3 = 130°

Finding values of angle x

x+x+130=180

2x=180-130

2x=50

x=25

So, Angle 1 = 25 °

Angle 2 = 25°

As, angle B is greater than 90°, the sum of angles added must be greater than 90°

So, Option A, C and D can't be correct, as their sum is less than 90°

Option B m<ZA+m<ZB = 155° is correct option.

Verify:

m<ZA = 25° (as found earlier)

m<ZB = 130° (given)

So, m<ZA+m<ZB = 155°

25°+130° = 155°

if a = 1,   and b = 2, then the answer is 47.99999999. if you round that to the nearest whole number, the answer is 48.

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step-by-step explanation: