Question

Solve the given triangles by finding the missing angle and other side lengths.

Answer 1

Answer:

1. a ≈ 27.65; B = 28°; c ≈ 19.17
2. a ≈ 163.30; B = 120°; c ≈ 59.77
3. A = 13°; b ≈ 0.3874; c ≈ 1.3737

Step-by-step explanation:

As long as you're seeking outside help, the use of a suitable tool is appropriate. Many graphing calculators, apps, and web sites are available for solving triangles.

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Here, you're given two angles and a side. The third angle is the difference from 180° of the sum of the other two. For example, in the first problem, ...

  B = 180° -A -C = 28°

The missing sides are found using the law of sines. Since you are given a side, use that as the reference, and its opposite angle as the reference angle. Then you have ...

  side = sin(opposite angle)×((reference side)/sin(reference angle))

Note that the factor in italics remains the same for both remaining sides.

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In the first problem, this becomes ...

  a = sin(112°)·14/sin(28°) ≈ 27.65

  c = sin(40°)·14/sin(28°) ≈ 19.17

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The remaining problems are worked in similar fashion.

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Comment on triangle solutions

These are straightforward because you're given two angles. If you're given two sides and an angle, the solution method varies depending on which angle you're given (opposite the long side, the short side, or the unknown side). In the last case, the law of cosines is involved. In the second case, there are likely two solutions. Once you have two angles, you can proceed as above.

Answer 2

Answer:

The answer to your question is below

Step-by-step explanation:

First picture

1.- Find B

The sum of the internal angles in a triangle equals 180°

    40° + 112° + B = 180°

    B = 180 - 40 - 112

    B = 28°

2.- Find c using law of sines

    $\frac{c}{sin C} = \frac{b}{sin B}$

    $c = \frac{b sinC}{sin B}$

    $c = \frac{14 sin 40}{sin 28}$

           c = 19.2

2.- Find a using law of sines

    $\frac{a}{sin A} = \frac{b}{sin B}$

    $\frac{a}{sin 112} = \frac{14}{sin 28}$

    $a = \frac{14 sin 112}{sin 28}$

    a = 27.65

Second picture

1.- Find B

The sum of the internal angles in a triangle equals 180

       A + B + C = 180

      B = 180 - 15 - 45

      B = 120°

2.- Find a

    $\frac{a}{sin45} = \frac{200}{sin 120}$

    $a = \frac{200 sin 45}{sin 120}$

           a = 163.3

3.- Find c

    $\frac{c}{sin 15} = \frac{163.3}{sin 45}$

    $c = \frac{163.3 sin 15}{sin 45}$

           c = 59.8

Picture 3

1.- Find A

The sum of the internal angles in a triangle equals 180°

    A + B + C = 180°

    A = 180 - 162 - 5

    A = 13°

2.- Find b

    $\frac{b}{sin 5} = \frac{1}{sin 13}$

    $b = \frac{1 sin 5}{sin 13}$

           b = 0.39

3.- Find c

      $\frac{c}{sin 162} = \frac{1}{sin 13}$

      $c = \frac{1 sin 162}{sin 13}$

             c = 1.37