Verified answer

If A and B have angle x, then you have,

x + x + (x+75) = 180

3x + 75 = 180

3x = 105

x = 35

Angles A and B are 35 degrees. C is 110 degrees.

Okay, we know the area of any triangle is 180 degrees.

The problem also tells you that angle C is 75 degrees larger than A and B, it also states that A and B have the same angle.

Angle A= x, Angle B = x and angle c = x+75

putting this all together.....x+x+(x+75) = 180 degrees

Now...3x + 75 =180

subtract 75 from both sides give us

3x=105

x = 35 degrees.

so now plug 35 degrees into your equations for A, B and C

A = 35 degrees, B = 35 degrees and C = 35 degrees +75 degrees = 110 degrees

Since m<A = m<B, we can let m<A = m<B = x. Then, since m<C is 75 degrees more than both m<A and m<B, we can let m<C = x + 75.

So:

m<A + m<B + m<C = 180

==> x + x + (x + 75) = 180

==> 3x + 75 = 180

==> 3x = 105

==> x = 35.

This gives m<A = m<B = 35 degrees and m<C = 35 + 75 = 110 degrees.

I hope this helps!

A+B+C =180.......Eq.1

As A=B=x(say) so Eq.1 becomes

x+x+C =180.....Eq.2

As C is 75 more than A so C=x+75 and therefore Eq.2 becomes

x+x+x+75=180

or

3x=180-75

or

3x=105 giving

x=105/3

...=35 degrees

Therefore

A=35°

B=35°

C=35+75=110°

Use the records you are able to desire to make simultaneous equations A + B + C = a hundred and eighty attitude A measures one hundred ranges below the sum of the measures of attitude B and attitude C so A - one hundred = B+C combine those 2 A + A-one hundred = a hundred and eighty 2A = 80 A = forty the measures of attitude C is forty ranges below two times the degree of attitude B C + forty = 2B B + C = one hundred forty Subtract those 2 equations forty - B = 2B - one hundred forty 3B = a hundred and eighty B = 60 C = 80

c=a+75

a=b

a+a+a+75=180

3a=105

a=35

b=35

c=110