If a triangle with the angles of 72 degrees, 62 degrees, and 46 degrees and one of the side lengths is 40, what would the length of the other two side be?

Answers

Answer 1

Answer:

The length of the other two sides are: 43.09 and 32.60

Step-by-step explanation:

Let the given side of the triangle be b(= 40), and the angles be;

A = [tex]72^{o}[/tex], B = [tex]62^{o}[/tex]and C = [tex]46^{o}[/tex].

Applying the Sine rule,

[tex]\frac{a}{SinA}[/tex] = [tex]\frac{b}{SinB}[/tex] = [tex]\frac{c}{Sin C}[/tex]

So that,

[tex]\frac{a}{SinA}[/tex] = [tex]\frac{b}{SinB}[/tex]

[tex]\frac{a}{Sin 72}[/tex] = [tex]\frac{40}{Sin62}[/tex]

a = [tex]\frac{0.9511*40}{0.8829}[/tex]

  = 43.0898

a = 43.09

Also,

[tex]\frac{b}{SinB}[/tex] = [tex]\frac{c}{Sin C}[/tex]

[tex]\frac{40}{Sin62}[/tex] = [tex]\frac{c}{Sin46}[/tex]

c = [tex]\frac{0.7193*40}{0.8829}[/tex]

  = 32.5881

c = 32.60

The length of the other two sides are: 43.09 and 32.60


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Answers

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