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Can someone PLZZZZ help
( solve triangles using the law of sines)

Can someone PLZZZZ help ( solve triangles using the law of sines)-example-1

2 Answers

7 votes

Answer:

10

Explanation:

31°/5 = 108°/AC

31°·AC = 108°·5

0.5·AC = 1·5

AC = 5/0.5

AC = 10

User Shmuels
by
5.3k points
4 votes

Answer:


AC=9.2

Explanation:

The Law of Sines states:


(sinA)/(a) =(sinB)/(b) =(sinC)/(c)

The lower case letters represent the sides opposite the angles: ∠A - side a, ∠B- side b, ∠C- side c. It helps to make a model.

To Find AC, or side b, we can use A and B because we have the appropriate information to solve for b. Insert values:


(sinA)/(a)=(sinB)/(b) \\\\(sin31)/(5)=(sin108)/(b)

Solve for b. Multiply both sides by b:


b*((sin31)/(5))=b*((sin108)/(b))\\\\b*(sin31)/(5) =sin108

Multiply both sides by 5:


5*(b*(sin31)/(5) )=5*(sin108)\\\\b*sin31=5*sin108

Divide both sides by sin 31:


(b*sin31)/(sin31) =(5*sin108)/(sin31) \\\\b=(5*sin108)/(sin31)

Enter the simplified equation into a calculator:


b=9.2

The length of AC is 9.2 units.

:Done

User Negin
by
5.5k points