Question Video: Using the Law of Sines to Calculate Unknown Lengths in a Triangle | Nagwa Question Video: Using the Law of Sines to Calculate Unknown Lengths in a Triangle | Nagwa

Question Video: Using the Law of Sines to Calculate Unknown Lengths in a Triangle Mathematics • Second Year of Secondary School

𝐴𝐵𝐶 is an obtuse triangle at 𝐴, where 𝑏 = 15 cm, tan 𝐶 = 6/5, and 𝑚∠𝐵 = 27°. Find the lengths 𝑎 and 𝑐, giving the answer to the nearest integer.

04:21

Video Transcript

𝐴𝐵𝐶 is an obtuse triangle at 𝐴, where 𝑏 equals 15 centimeters, tan of 𝐶 equals six over five, and the measure of angle 𝐵 equals 27 degrees. Find the lengths 𝑎 and 𝑐, giving the answer to the nearest integer.

It’s always a good idea to begin by sketching a diagram. This doesn’t need to be to scale, but it should be roughly in proportion. So we can check the suitability of any answers we get. We’ll label the triangle such that the obtuse angle is angle 𝐴. We’ll then add the other information we know. Since we’ve been told that the tan of 𝐶 equals six-fifths, we can solve this equation to find the angle at 𝐶. The inverse tan of the tan of 𝐶 equals 𝐶. And the inverse tan of six-fifths is equal to 50.194 continuing degrees. Since we know we’re going to be rounding in a later step, it’s best not to round at this point. This will prevent errors by rounding too early.

Next, we can solve for the measure of the angle at 𝐴. We know that the angles in a triangle sum to 180. Therefore, the measure of angle 𝐴 will be equal to 180 degrees minus 27 degrees plus 50.194 continuing. Solving gives us 102.805 continuing degrees. This angle is an obtuse angle; it is greater than 90 degrees. Now, we have a non-right triangle. And we know the lengths of one side and all the angles. This means we can use the law of sines to find any missing measurement. We don’t use the law of cosines here as that requires a minimum of two known sides.

The formula for the law of sines is 𝑎 over sin 𝐴 equals 𝑏 over sin 𝐵, which equals 𝑐 over sin 𝐶. Alternatively, this can be written as sin 𝐴 over 𝑎 equals sin 𝐵 over 𝑏, which equals sin 𝐶 over 𝑐. Since we’re trying to solve for side length, we’ll use the first form. Either form will work, but by using the first form, we reduce the amount of rearranging we’ll need to do to solve. Before we move on, let’s label the sides of our triangle. The side opposite angle 𝐴 will be labeled with a lowercase 𝑎. The side opposite angle 𝐵 will be lowercase 𝑏. And the side opposite angle 𝐶 will be lowercase 𝑐.

First, let’s calculate the length of side 𝑎. This means we’ll need 𝑎 over sin 𝐴 equals 𝑏 over sin 𝐵. We’re using 𝑏 over sin 𝐵 as we know both value 𝑏 and the angle at 𝐵. Substituting what we know, we get 𝑎 over sin of 102.805 continuing is equal to 15 over sin of 27 degrees. Once again, we’ll use the exact value for the measure at angle 𝐴. To solve, we’ll multiply both sides of the equation by sin of 102.8 continuing. 𝑎 is therefore equal to 15 over sin of 27 degrees times sin of 102.8 continuing degrees. Popping that into the calculator gives us 𝑎 equals 32.2185 continuing. That value to the nearest integer will be 32 centimeters.

We’ll follow the same process to solve for the length of side 𝑐. Using 𝑐 over sin 𝐶 is equal to 𝑏 over sin 𝐵. Substituting what we know, we’ll have 𝑐 over sin of 50.194 continuing degrees is equal to 15 over sin of 27 degrees. Then we’ll multiply through by sin of 50.194 continuing degrees. 𝑐 equals 15 over sin of 27 degrees times sin of 50.194 continuing degrees, which gives us 25.382 continuing. And to the nearest integer, side 𝑐 is then 25 centimeters.

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