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

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. Now we notice we have a triangle here with two missing side links and one missing angle. And four possible answers with the missing values side P Q side pr and angle P. Let's see if we can find our missing values. The first thing we can do is find our final missing angle. We know the sum of interior angles of a triangle equals degrees. We can add together all of our interior angles and instead it equals to 1 80 where we have angle P plus angle Q plus angle R, 180 degrees will then equal angle P which is missing plus 40 degrees plus 44 degrees. We can solve for ANG LP by subtracting Q and R on both sides, we should get ang LP as equals to 96 degrees. Now that we have all our angles, we can make use of the law of signs to find our missing sets the law of science sets up three ratios where our side links is divided by the sign of the angle opposite of the side. For example, let's say we're looking for side pr we can write this as pr by the best sign of its opposite angle, the opposite angle will just take us from the side across the triangle which in our case are corresponding with the angle. Here is angle Q 40 degrees will do the same for every triangle. We also have side QR defined by its opposite angle. P will say sine of 96 degrees because it just found what he was. We have side P Q which is divided by the sign of angle R. So sine at 44 degrees. Now, we also know what side QR is, which is eight units. Now let's go ahead and solve for Pr and P Q, we have pr divided by sine 40. It equals to eight divided by sine 96 because that's the only side that has an actual measure. Let's go ahead and solve for Pr we multiplied both sides by sine 40 to get pr equals eight sine 40 divided by sine 96. We get pr it's approximately 5.2 units. If we throw this into a calculator, let's do the same or other side. P Q, we have P Q divided by sine 44 degrees. S equals to eight divided by a sun of 96 degrees. Multiplying both sides by sine 44. Gives us P Q as approximately eight sign 44 divided by a sign of P Q then approximates to be about 5. units. We now have our angle P of 96 degrees five pr or R P of 5.2 units inside P Q of 5.6 units. If we look at our possible answers, we see the answer to our problem is answer A 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

Types of Triangles and Their Properties

Law of Sines and Law of Cosines

Rounding and Measurement Accuracy

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