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

Whatever the triangle, the sum of the 3 angles is always 180 °.

First triangle

You can write:

68 + 56 + C = 180

124 + C = 180

C = 180 - 124

C = 56

As the angle C is equal to the angle A, you can deduce that the triangle is an isosceles triangle.

As the triangle is an isosceles triangle, you can deduce that: BA = BC.

AB = 1.3

Then you can see that:

AC.cos(A) + BC.cos(C) = AC → we've just seen that: C = A = 56

AC.cos(56) + BC.cos(56) = AC

AC - AC.cos(56) = BC.cos(56)

AC.[1 - cos(56)] = BC.cos(56)

AC = BC.cos(56) / [1 - cos(56)] → given that: BC = 1.3

AC = 1.3 * cos(56) / [1 - cos(56)] → where: cos(56) ≈ 0.55919

AC ≈ 1.6491135

AC = 1.65

Second triangle

You can write:

15 + 45 + B = 180

60 + B = 180

B = 180 - 60

B = 120

As the angle C is equal to the angle A, you can deduce that the triangle is an isosceles triangle.

As the triangle is an isosceles triangle, you can deduce that: BA = BC.

AB = 1.3

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

15 + 90 + yellow = 180

105 + yellow = 180

yellow = 180 - 105

yellow = 75

You can write:

45 + 90 + green = 180

135 + green = 180

green = 180 - 135

green = 45

You can see:

AB.cos(green) = BC.cos(yellow)

AB.cos(45) = BC.cos(75)

You can see:

BC.cos(15) + AB.cos(45) = 200 → recall the previous result: AB.cos(45) = BC.cos(75)

BC.cos(15) + BC.cos(75) = 200

BC.[cos(15) + cos(75)] = 200

BC = 200/[cos(15) + cos(75)] → where: cos(15) ≈ 0.96592 and where: cos(75) ≈ 0.25881

BC ≈ 163.299

BC = 163.3

Recall: AB.cos(45) = BC.cos(75)

AB = BC.cos(75) / cos(45) → recall BC

AB = { 200/[cos(15) + cos(75)] }.cos(75) / cos(45)

AB = 200 * cos(75)/{ cos(45).[cos(15) + cos(75)] }

AB ≈ 59.77169

AB ≈ 59.78

C = 180 - 68 - 56 = 56

BA = 1.3 (the triangle is isosceles

... law of cosines

CA^2 = 1.3^2 + 1.3^2 - 2*1.3*1.3cos68

CA = 1.454

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now you try the others

Link1

1/

<C = 180-(68+56)=56

Sine rule --> 1.5/sin56 = AB/sin56 = AC/sin68

easy to find AB and AC

2/ similar 1/

<B=120

200/sin120 = AB/sin15 = BC/sin45

solve for AB and BC