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 and angle of the triangle. P. QR express the links to one decimal place and the angle to the nearest degree that we noticed here, we have a triangle P. QR with some missing values. And we have four possible answers. With those missing values, two of them are sides and one of them is an angle. Let's see if we can figure out what we're missing. Now, the first thing we can do is we can find our final angle. We know this, we have angle R and angle Q but we don't have angle P. However, we can solve for angle P in the triangle, the sum of the interior angles will be equal to degrees. In other words, we can say angle P plus angle Q plus angle R should equal 108 degrees. We have angles Q and R. So we just need to find angle P Q is degrees and R is 106 degrees. We can set that equals to 1 80 solve for P, we should get angle P, it equals to 42 degrees whenever we subtract our Q and our R to the other side. Now that we have our angle P, we can make use of the law of science to determine our missing side links. A lot of sign states, we have three ratios of our triangle and these ratios are based on our sides and our ankles. For instance, our angle R has an opposite side of P Q. If we're to draw a line for angle R directly to the other side of the triangle. In law of science, this ratio is our side length across from the angle divided by the sign of the angle. This equals to our other side in angle ratios. For instance, we can have side QR which is opposite from angle P. In our case, our P was 42 degrees. We also have the side pr which is opposite angle Q. Now we have side linked pr which is 30. We cannot make use of these ratios to find our missing side links. Let's first say P Q divided by sine one oh six is equals to 30 divided by sine 32 degrees. Solving for P Q means we multiply a sign of 100 and six degrees. On both sides, we get P Q equals 30 sine 106 divided by sine 32. If you throw that into a calculator, that approximates to about 54.4 units. Now we can do the same for side QR, we have QR divided by a sine, 42 degrees equals 30 divided by sine 32 degrees, which gives us QR equals sine 42 degrees divided by a sine 32 degrees. We get QR is approximately 37.9 units. We now have our missing side links and our missing angle pay 42 degrees, 54.4 units and 37.9 units. If we look at our possible answers, we see answer. D is the answer to our problem? 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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