Please help me with my Trigonometry?

1.

Solve for all missing sides and angle

Triangle ABC AB = 12 Angle B = 45° Angle C = 52°

Angle A = 83°

Acute Scalene Triangle

Side a = 15.11471

Side b = 10.76798

2.

Solve for all missing sides and angle

Triangle ABC Angle A = 45° Angle C = 15° AC =200

Angle B = 120°

Obtuse Scalene Triangle

Side a = 163.29932

Side c = 59.7717

1.

A + B + C = 180° (∠ sum of Δ)

A + 45° + 52° = 180°

A = 83°

BC / sinA = AB / sinC (sine law)

BC / sin83° = 12 / sin52°

BC = 12 × sin83° / sin52°

BC = 15.1

AC / sinB = AB / sinC (sine law)

AC / sin45° = 12 / sin52°

AC = 12 × sin45° / sin52°

AC = 10.8

====

2.

A + B + C = 180° (∠ sum of Δ)

45° + B + 15° = 180°

B = 120°

AB / sinC = AC / sinB (sine law)

AB / sin15° = 200 / sin120°

AB = 200 × sin15° / sin120

AB = 59.8

BC / sinA = AC / sinB (sine law)

BC / sin45° = 200 / sin120°

BC = 200 × sin45° / sin120°

BC = 163.3

Like the other time, whatever the triangle, the sum of the 3 angles is always 180 °.

You can write:

A + 45 + 52 = 180

A + 97 = 180

A = 83

Then you draw the red line (perpendicular to the line (BC), and you can write:

AM = 12.sin(45)

AM = AC.sin(52)

AC.sin(52) = 12.sin(45)

AC = 12.sin(45) / sin(52) → recall: sin(45) = (√2)/2

AC = 12.[(√2)/2] / sin(52)

AC = (6√2) / sin(52)

AC ≈ 10.7679

AC = 10.77

BC = AC.cos(52) + AB.cos(45) → recall AC

BC = [(6√2) / sin(52)] + AB.cos(45) → you know that: AB = 12

BC = [(6√2) / sin(52)] + 12.cos(45) → recall: cos(45) = (√2)/2

BC = [(6√2) / sin(52)] + 12.[(√2)/2]

BC = [(6√2) / sin(52)] + 6√2

BC ≈ 19.2532

BC = 19.26

In both cases, you are given a triangle with two known angles and one known side.

The sum of the angles of any triangle is 180°. Use that fact to find the one unknown angle.

The law of sines states that in any triangle, the sines of the angles are proportional to the lengths of their opposite sides. Use that to find the two unknown sides.