# Question Video: Using the Law of Sines to Calculate All Possible Measures of an Angle in a Triangle Mathematics

For a triangle 𝐴𝐵𝐶, 𝑎 = 2 cm, 𝑏 = 5 cm, and 𝑚∠𝐵 = 40°. Find all possible measures of ∠𝐴 to the nearest degree.

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### Video Transcript

For a triangle 𝐴𝐵𝐶, 𝑎 is equal to two centimeters, 𝑏 is equal to five centimeters, and the measure of angle 𝐵 is 40 degrees. Find all possible measures of angle 𝐴 to the nearest degree.

In this question, we are given the lengths of two sides of our triangle together with a measure of one of the angles. And we are asked to find all the possible measures of a second angle 𝐴. If triangle 𝐴𝐵𝐶 exists from the measures given, we know that we can find one possible measure of angle 𝐴 using the law of sines. This states that sin 𝐴 over 𝑎 is equal to sin 𝐵 over 𝑏, which is equal to sin 𝐶 over c, where uppercase 𝐴, 𝐵, and 𝐶 are the measures of the three angles and lowercase 𝑎, 𝑏, and c are the lengths of the sides opposite the corresponding angles.

Substituting the values we’re given in this question into the law of sines gives us sin 𝐴 over two is equal to sin of 40 degrees over five. We can multiply through by two such that sin 𝐴 is equal to two multiplied by sin 40 degrees all divided by five. Typing this into our calculator gives us 0.2571 and so on. We can then take the inverse sine of both sides of this equation, giving us 𝐴 is equal to 14.89 and so on. We are asked to give our answer to the nearest degree. So one possible measure of angle 𝐴 is 15 degrees.

To work out whether there is a second possible solution, we recall that the sin of 180 degrees minus 𝜃 is equal to sin 𝜃. As 180 degrees minus 15 degrees is 165 degrees, then the sin of 165 degrees must equal the sin of 15 degrees. This means that we need to check whether 165 degrees is a valid solution. Since angles in a triangle sum to 180 degrees and 165 plus 40 is equal to 205, 165 degrees is not a valid solution to this problem. We can therefore conclude that the only possible measure of angle 𝐴 is 15 degrees to the nearest degree.

It is worth noting that we could’ve established that there was only one possible solution from the measurements given in the question, since if angle 𝐵 is acute and side length 𝑏 is greater than side length 𝑎, then only one triangle exists. And this means that there is only one possible measure of angle 𝐴.