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 in angle of the triangle. P. QR express the links to one decimal place and the angle to the nearest degree where we have a triangle P. QR with a couple angles. In one side, our possible answers are two missing sides, QR R P and one missing angle R. Let's see if we can find all these missing values. The first thing we notice is that we can find the angle R, we know that the sum of the interior angles of a triangle should equal 180 degrees. In other words, angle P plus angle Q plus angle R equals 180 degrees. We can substitute because we know angles P and Q as eight degrees and 16 degrees. We don't have A R. So we want to solve for that, we can subtract our eight and our 16. On the other side, we then get angle R equals 156 degrees. Now we have our angle, we can find our missing sites to do this. We'll make use of the law of sites. This tells us the side opposite from our angle can be written as a ratio. For example, let's take side, QR QR will be divided by our opposite angle, which if we draw a line directly to the other side of the triangle, we have angle P. If other word, our denominator will be sign of the angle P, we do the same for the other sites. Pr for example, is divided by the sign of the opposite angle Q 16 degrees. And finally, we have P Q which has the opposite angle R than a sign of 156 degrees. Now we have our P Q length, which we see is 80 units. Let's use our ratios to find QR and Pr, we have QR divided by sine eight degrees. It's equals to 80 divided by a sine of 156 degrees. If we solve for QR, we would get QR equals 80 multiplied by sine eight degrees divided by a sine of 1 56 degrees multiplying our sine of eight degrees over to the other side. Now we can just throw this part into a calculator but we get QR approximates to be about 27.4. We can do the same for pr, we have pr divided by a sine of 16 degrees as equals to 80 divided by the sine of 156 degrees, which can be rewritten as Pr equals sign 16, divided by a sine of 156 degrees. Throw that into a calculator to get pr approximates to 54.2 units with the units for QR as well. We have our angle R which is 100 and 56 degrees side. QR at 27.4 units and pr at 54.2 units which corresponds to answer B 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

Law of Cosines

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