Question

The point (5, −2) is on the terminal ray of angle θ, which is in standard position. Without evaluating, explain how you would find the values of the six trigonometric functions.

Answer 1

1. Check the picture attached.

let P, M, N be the points :P(5, -2), M(5, 0), N(0, -2)

ABP is the angle with measure ∅, where BA is the initial ray and BP the terminal ray.

2. m(PNM)=∅ as can be easily checked

In triangle PNM, NP=5, PM=2 and NM=$\sqrt{ 5^{2} + 2^{2} }= \sqrt{25+4}= \sqrt{29}$ units


3. let opp=opposite side, adj=adjecent side, hyp=hypothenuse

sin ∅ = opp/hyp=$\frac{2}{\sqrt{29}}$

cos ∅ = adj/hyp=$\frac{5}{\sqrt{29}}$

tan ∅ = opp/adj=2/5

cot ∅ = adj/opp=5/2

sec ∅ = hyp/adj=$\frac{\sqrt{29}}{5}$

csc ∅ = hyp/opp=$\frac{\sqrt{29}}{2}$

Answer 2

Answer:

below

Step-by-step explanation:

⇒Position of point , which is on the terminal ray of Angle Theta = (5, -2)

⇒Let this point be Represented as P (5, -2) and Origin is Represented by Point O(0,0).

⇒Join OP and draw Perpendicular from P on the X axis,which cut the X axis at A(5,0).

This is Required Right Δ O AP.

⇒Find the Length of OP , using Distance formula.

O A=5 units

PA=2 units

⇒As, the point lies in fourth Quadrant,Only Secant and Cosine function will be positive, all other trigonometric function will have negative value.

By Applying , trigonometric ratio formula,as we have length of three sides of triangle we can evaluate six trigonometric ratios.