In Exercises 1–8, solve each triangle. Round lengths of sides to ...

Question

In Exercises 1–8, solve each triangle. Round lengths of sides to the nearest tenth and angle measures to the nearest degree.

Accepted Answer

Find the missing side links to an angle of the triangle. P. QR express the links to one decimal place and the angle to the nearest degree. Now we noticed we have a triangle with two missing sides and one angle. And our possible answers are four different answers with our two missing sides. And our angle, in particular, we have sites P Q QR and angle R unknown. Let's see if we can find what these missing values are. First, we can find our angle. We know the sum of interior angles of a triangle should equal degrees. Let's say we have angle P plus angle, Q plus angle R equals 180 degrees. Now we have angles P and Q 1 16 degrees and 29 degrees. You will add angle R and set it equals to 1 80. Now we can solve for angle R subtracting 1 16 and 29 from both sides, we get angle R is equals to 35 degrees. Now that we have all of our angles, we can make use of the law of science. The love signs tells us that we can set up a ratio as long as we have the side length. And this opposing angle, for example, we have side link 13 for side pr we can write this as pr divided by a sign of its opposite angle which in our case, if we go directly across the triangle, our angle is angle Q. So we will say that is sine divided by the angle of degrees. We'll do the same for all of our other sites. We have QR divided by its opposite angle, angle P or sine 1 16 degrees. We have P Q divided by a sine of the angle which is angle R sign of 35 degrees which we just found. Now, he knows what side pr is that is 13 units. Now that we have three ratios we can solve for our missing sets. We first have 13 divided by a sine 29 degrees equals QR divided by a sine of 1 16 degrees. We can multiply both sides by sine 1 16 to get QR by itself. QR equals 13 sine 1 divided by 9 29. If we were to approximate this in a calculator outside late QR would be approximately 24. units. We can do the same for P Q P Q divided by sine 35 equals to 13 divided by a sign 29. We will get P Q, it equals the 13 sine 35 degrees divided by sine 29 degrees dis approximates to be about 15.4 units. We now have all of our missing sides and angle angle R is 35 degrees side P Q is 15. units and side QR is 24.1 units. If we look at our possible answers, we determine the answer to our problem is answer C OK. I hope to help you solve the problem. Thank you for watching. Goodbye.

In Exercises 1–8, solve each triangle. Round lengths of sides to the nearest tenth and angle measures to the nearest degree.

Key Concepts

Triangle Types

Law of Sines and Law of Cosines

Rounding and Precision

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