12. If DH = HE, MBG = (9x – 20) and
MGC = (5x + 28), find mAB.

Given:

DH = HE

m(ar BG) = (9x - 20)°

m(ar GC) = (5x + 28)°

To find:

The m(ar AB)

Solution:

Radius bisects the chord and its arc.

m(ar BG) = m(ar GC)

(9x - 20)° = (5x + 28)°

9x° - 20° = 5x° + 28°

Add 20° on both sides.

9x° - 20° + 20° = 5x° + 28° + 20°

9x° = 5x° + 48°

Subtract 5x° from both sides.

9x° - 5x° = 5x° + 48° - 5x°

4x° = 48°

Divide by 4 on both sides, we get

x° = 12°

Substitute x = 12 in m(ar BG) and m(ar GC).

m(ar BC) = m(ar BG) + m(ar GC)

              = (9x - 20)° + (5x + 28)°

              = (9(12) - 20)° + (5(12) + 28)°

              = 88° + 88°

m(ar BC) = 176°

Chord AB congruent to chord BC.

m(ar AB) = m(ar BC)

m(ar AB) = 176°

The measure of arc AB is 176°.

50.6

step-by-step explanation:

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