Answer:
E. 0.759
Step-by-step explanation:
You can take this right triangle to have a base length side LM , height of LN and a hypotenuse of MN
The sine of angle ∠LMN is 0.759, find the value of ∠LMN

∠LMN=49.38°
Find angle ∠LNM
You know sum of angles in a triangle add up to 180°, given that this is a right-triangle, the base angle is 90° hence
∠LNM=180°-(90°+49.38°)
∠LNM= 180°-139.38°=40.62°
Find cos 40.62°
Cos 40.62°=0.7590
a. The
component tells you the particle's height:

b. The particle's velocity is obtained by differentiating its position function:

so that its velocity at time
is

c. The tangent to
at
is

Use this version of the Law of Cosines to find side b:
b^2 = a^2 + c^2 − 2ac cos(B)
We want side b.
b^2 = (41)^2 + (20)^2 - 2(41)(20)cos(36°)
After finding b, you can use the Law of Sines to find angles A and C or use other forms of the Law of Cosines to find angles A and C.
Try it....