The point (0.28, 0.96) lies on the unit circle. What is the tangent of an angle drawn in standard position whose terminal ray passes through this point?

Step-by-step explanation:

Given that the point that lie on the circle is

P = (0.28, 0.96)

Check attachment for understanding of the diagram

Generally

P = (sinθ, Cosθ)

Therefore, compare this with the given data,

Sinθ = 0.28

Cosθ = 0.96

So, the tangent drawn will be

Tan θ = opposite / adjacent

Tan θ = Sinθ / Cosθ

Tan θ = 0.28 / 0.96

Tan θ = 0.2917

θ = arcTan(0.2917)

θ = 16.26°

The tangent of the angle drawn is 16.26°

You can as well use the trigonometry formula,

So, from the attachment,

Tan θ = opposite / adjacent

Tan θ = 0.28 / 0.96

Tan θ = 0.2917

θ = arcTan(0.2917)

θ = 16.26°

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