In triangle ABC, the angle C is six times as large as angle A. The measure of angle B is 52 greater than that of angle A. How do you find the measure of the angles?
A = 16, B = 68, C = 96
I'll try to explain this the best I can, we know that all triangles angles have to have a sum equal to 180 degrees. Therefore, we can create the equation A + B + C = 180. We also know that B is 52 degrees greater than angle A, so we can create the equation B = A + 52. We also know that C is 6 times larger than angle A, so we can create the equation C = 6 x A. We can them substitute these two equations into the first equation, getting A + (A + 52) + (6A) = 180. This can be simplified to 8A + 52 = 180, which then we can subtract the 52 from the 180, and divide the 8, getting that A = 16.
From here we can either solve for B or for C, I will explain B first. We can take our original B equation, B = A +52, and plug in A as 16, for which we get B = 68. We can then take two ways of solving for C, since we know both B and A, we can do the original equation, A + B + C = 180, or we can use C = 6A. Using the original equation, we get 16 + 68 + C = 180, which is then equated to C = 96. We can check it using the other equation, C = 6A, which is C = 6(16), which is confirmed to be 96.
I hope this was helpful.