in adjoining figure AE||BC, ?DAE = x, ?ACB = (x-15), ?BAC = (x/2 +y) and ?ABC = (y+15). find the value of x and y.

Question

in adjoining figure AE||BC, ∠DAE = x, ∠ACB = (x-15), ∠BAC = (x/2 +y) and ∠ABC = (y+15). find the value of x and y.

Hints:   i) DAB is a striaght angle and ∠CAE = ∠ACB = (x - 15)

           ii) Sum of triangle ABC = 180

Answer 1

The measures of the angles of triangle ABC add up to 180 degrees:

 (x-15) + (x/2 +y)  + (y+15) = 180

(x + x/2) + (y + y) + (-15 + 15) = 180        Like terms grouped together

3x/2 + 2y = 180

We also know that ∠DAB = x + (x - 15) + (x/2 + y) and that ∠DAB is a straight angle, so:

x + (x - 15) + (x/2 + y) = 180

5x/2 + y -15 = 180

5x/2 + y = 195

We now have a system of equations we can use to solve for x and y:

5x/2 + y = 195
3x/2 + 2y = 180

There are multiple ways to solve a system of equations; I chose to multiply the top equation by -2 and use elimination:

-5x - 2y = -390
3x/2 + 2y = 180  
-7x/2     = -210

Multiply both sides by -2/7 and we get:

x = 60

You can plug that in to the equation of your choice and find that y = 45.

Just to check, we can find that the angles of the triangle are 45, 75 and 60, which add to 180.

Answer 2

We do not see the figure.   Yet we know line AE is parallel to line BC.

THerefore, angle ABC must equal 90 degrees.  

From algebra we can figure from the given facts the values of X and Y.

90 = X + 15  .= Y - 15.

SO, X = 105 degrees and Y = 75 degrees