We were given a refresher packet for Trigonometry in math today and I got all question but three. I tried them but keep getting the incorrect answer. Can somebody please explain? Thank you!

Question

1. Given that sinA=1/3 and in Quadrant 1 and cos B= -3/5 and in Quadrant 2.

a.) State cos A and sin B

b.) Find exactly sin (A+B)

2. Solve for all possible triangles ABC such that A=63 degrees, a=15 cm and c=16 cm. (For this one I was told that I didn't have ALL of the solutions)

3. Given a triangle ABC with <B=39 degrees and side a=9 cm. What are the possible values for the length of side b so that there is at least one triangle ABC?

Thank you

Answer 1

1.a, cosA in a triangle is the length from A to the point where the altitude from C to side c is divided by the side b (the "hypotenuse") sinB in a triangle is the length of the altitude from C to side c is divided by the side a (the "hypotenuse")

1.b. sin(a+B)=sinA*cosB+sinB*cosA

2.This is the ambiguious case, Side Side Angle (SSA). Use the law of cosine

and put into quadradic form to solve for b values that satisfy the equation.

b²-2bc*cosA+(c²-a²)=0,,,,, if the discriminant (4*c²*cos²A-4*(c²-a²))>0 then there are two values for b (and angles B and C), if the discriminant =0 then the is only one triangle possible, if the discriminant is <0 then there is no triangle possible.

3.Using law of sines for triangle, sinB/b=SinA/a,,,, SinA=(a*sinB)/b,,, I know SinA≤1

which implies b≥a*sinB

using my calculator, for problem 2. I fonud b at either 12 or 2.6. you can use the law of sines to compute the angles B and C for each b). Also using my calculator, for problem 3. b≥5.8.