In a triangle, one angle is 9 degrees less than twice another. The third angle is 18 degrees more than twice their sum. Find the angles.

Answer 1

FIRST OF ALL, IN ANY GIVEN TRIANGLE, THE THREE INTERIOR ANGLES WILL ALWAYS ADD UP TO 180.

So, we can start with this: Let a, b, and c represent the 3 angles. If a, b, and c are the interior angles of a triangle, then

a + b + c = 180

∠a is 9 degrees less than twice angle ∠b

So, ∠a = 2b - 9

∠c = 2(a + b) + 18

C = 2(a + b) + 18, and if a = 2b - 9, then we can substitute (2b - 9) for a in the previous equation for ∠c

C = 2((2b - 9 + b)) + 18

C = 4b - 18 + 2b + 18

C = 6b

Now we can substitute the new expressions which represent angles a and c into the original equation at the top.

(2b - 9) + b + 6b = 180

9b - 9 = 180

+ 9 + 9

9b = 189

÷ 9 ÷ 9

b = 21, so the measure of angle b is 21 degrees.

Now we know that angle a = 2b - 9, lets plug in the value of b to find a.

a = 2(21) - 9 = 42 - 9 = 33 degrees

and the measure of angle c = 6(21) = 126 degrees

Answer 2

A=2B-9

C= 18+ 2(A+B)= 18+4B-18+2B= 6B

A+B+C = 180

2B-9 + B + 6B = 180

9B = 189

B= 21 degrees

A= 2(21)-9= 33 degrees

C= 180-33-21= 180-54= 126 degrees

C=18+2(33+21) = 18+2(54)= 18+108 = 126

the 3 angles are 21, 33 and 126 degrees