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Determine the measure of angle ABC to the nearest degree

Determine the measure of angle ABC to the nearest degree-example-1

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Answer:

mABC = 59°

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°

User KillaKem
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