Determine the measure of angle ABC to the nearest degree

mABC = 59°

Step-by-step explanation:

To find the value of x, we can call the height BD by 'h' and then use the Pythagoras' theorem in the triangles ABD and BCD:

triangle ABD:

(3x + 1)^2 = (x + 1)^2 + h2

9x2 + 6x + 1 = x2 + 2x + 1 + h2

h2 = 8x2 + 4x

triangle BCD:

(4x - 1)^2 = (2x + 1)^2 + h2

16x2 - 8x + 1 = 4x2 + 4x + 1 + 8x2 + 4x

4x2 - 16x = 0

x2 - 4x = 0

x - 4 = 0

x = 4 units

Now we can find the angles mABD and mBDC using the sine relation:

sin(mABD) = (x + 1) / (3x + 1) = 5 / 13

mABD = 22.62°

sin(mBDC) = (2x + 1) / (4x - 1) = 9 / 15

mBDC = 36.87°

So we have that:

mABC = mABD + mBDC = 22.62° + 36.87° = 59.49°

Rounding to the nearest degree, we have mABC = 59°

this doesn't make sense lol

step-by-step explanation:

we are given

since, we have to isolate x on anyone side

so, we can move all x terms on one side

subtract both sides by 2x

add both sides by 9

divide both sides by 4

we get