please help

Question

the point (-2,7)  is located on the terminal arm of<A in standard position.

a) what are the exact primary trigonometric ratios for <A?

b) use a calculator to determine the measures of <A and <B to the nearest degree.

Answer 1

(a).

The point  (-2, 7).

Let A be an angle in standard position with (x, y) a point on the terminal side of A and r = √(x² + y²) ≠ 0.

To find the value of r, substitute x = - 2 and y = 7 in the above formula, we obtain

r = √[(2)² + (7)²]

^{r = √[4 + 49]}

r = √53.

Therefore,

sin(A) = y/r = 7/√53.

cos(A) = x/r = - 2/√53.

tan(A) = y/x = 7/(- 2) = - 7/2.

cot(A) = x/y = (- 2)/(7) = - 2/7.

sec(A) = r/x = √53/(- 2) = - √53/2.

csc(A) = r/y = √53/(7) = √53/7.

(b).

The value of sin(A) = - 7/2 -----> A = sin^-1(7/√53) ≅ 74.0546^o = 74.

One angle C = 90 and remaining angle is B.

The total of three angles in a triangle is 180^o.

A + B + C = 180^o

74 + B + 90 = 180^o.

B + 164 = 180^o.

B = 16^o.