Question 1206928
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The Law of Sines is a bit overkill, but it is a method you could follow.


A more easier approach would be to use the SOH-CAH-TOA trig ratios<ul><li>SOH = sine opposite hypotenuse</li><li>CAH = cosine adjacent hypotenuse</li><li>TOA = tangent opposite adjacent</li></ul>I'll focus on problem 1 only and leave the others for the student.


C = 90 degrees
A = 15 degrees
B = 90-A = 90-15 = 75 degrees
Angles A and B are complementary, i.e. they add to 90 degrees, since we're dealing with a right triangle.


Now that we know the angles, we can determine the side lengths.
sin(angle) = opposite/hypotenuse
sin(A) = a/c
sin(15) = a/37
a = 37*sin(15)
a = 9.576305 approximately
Please make sure that your calculator is set to degrees mode. 
I often check by typing in something like "sin(30)" and the output for this example should be 0.5


And,
sin(angle) = opposite/hypotenuse
sin(B) = b/c
sin(75) = b/37
b = 37*sin(75)
b = 35.739256 approximately


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Here is the fully solved triangle for problem 1
Angles: A = 15, B = 75, C = 90
Sides: a = 9.576305 approximately, b = 35.739256 approximately, c = 37
Round the approximate values however needed.


I'll leave the sketch for the student to do, as well as problems 2 through 5.


More practice found here
<a href="https://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometry-basics.faq.question.1202430.html">https://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometry-basics.faq.question.1202430.html</a>
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