In a pair of complementary angles, the larger angle is six more than three times the smaller angle. the measures of the angles are

2 angles are Complementary, if the sum of their measures = 90°.

Let the 2 angles be A and B, and let their measures be m(A)=a, m(B)=b.

Let A be the larger angle.

"the larger angle is six more than three times the smaller angle."

can be 'translated' into algebra as:   a = 6 + 3b

Also, since A and B are complementary, we have a+b=90°.

substitute a=6+3b in the second equation:

a+b=90°

(6°+3b+b)=90°

4b=90°-6°

b=84°/4=21°

Then, a = 90°-21°=69°

a= 69°, b=21°

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12