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in the figure, m
bm4c and 2d is a right angle.
cos b=
sin a
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sin d
o

sin (A)

Explanation:

In the given figure the "hidden" vertex of the larger triangle is A.

Since, angle D is a right angle, both triangles BDC and CDE are right triangles (and they are also similar).

Since angle B and angle C have the same measure and angle D measures 90° (right angle), angle A and angle C are complementary, i.e.

∠ C + ∠ A = 90°

For complementary angles, the sine of one acute angle equals the cosine of the other acute angle.

So, the sine of angle C is equal to the cosine of angle  B, i.e. cos (B) = sin (A).

The correct answer would be the function g(x) increased at a faster rate

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