# 30 30. Prove that the sum of three angles of a triangle is 180$\text{°}$. Using this result, find the value of x and all three angles of triangle, it the angles are (2x + 7)$\text{°}$ , (2x- 25)$\text{°}$and (3 -12)$\text{°}$.

Draw a line PQ through the vertex A and parallel to the side BC of a triangle ABC. Now, angle PAB + angle BAC + angle CAQ = 180 degrees (as PQ is a straight line). Equation 1. Angle PAB = angle ABC (as PQ AND BC parallel lines and AB is transversal ;alternate interior angles). ] Equation 2 Angle CAQ = angle ACB (as PQ AND BC parallel lines and AC is transversal ;alternate interior angles). Equation 3 Substitute angle PAB and angle CAQ in Equation 1 by angle ABC and angle ACB (as found in Equation 2 and Equation 3) Thus we get, Angle ABC + angle BAC + angle ACB = 180 degrees. In other words, in the triangle ABC, angle B + angle A + angle C = 180 degrees. Thus the sum of all the angles of a triangle is 180 degrees Proved.
• 3
Since the first part has been answered, I will answer the second part.
2x+7+2x-25+3x-12=180
7x-30=180
7x=210
x=30.
So,
2x+7=67,
2x-25=35.
3x-12=78.
Hope it helps.
• 0
all are angle are = 180 then solve it
• -2
most simple

• -1
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