In Exercises 9–24, solve each triangle. Round lengths to the near...

Question

In Exercises 9–24, solve each triangle. Round lengths to the nearest tenth and angle measures to the nearest degree.a = 5, b = 7, c = 10

Accepted Answer

Find the missing side links. And angles of the triangle express the links to one decimal place in the angle to the nearest degree where we have P equals six Q equals nine or equals 13. And all of our answers have three different angles. PQ and R not to solve this problem, we will make use of the law of cosine la cosines tells us that we have a side A squared which is equals to B squared plus C squared minus two BC multiplied by cosine of angle. A similarly, we'd have the same for B squared which is A squared plus C squared minus two AC cosine angle B and finally C squared as equals to A squared plus B squared minus two A B cosine of C. We can modify all of these using our sides PQ. And R let's say P is first PS it equals to Q squared plus R squared minus two QR cosine angle P. Now we want the angle so we will solve repeat, we get six squared is equals to nine square plus 13 squared minus two, multiply it by nine and cosign pete, you get six squared minus nine squared minus 13 squared as equals to negative two multiplied by nine and cosine peak. No, we can go ahead and simplify all of our things here. 36 minus 81 minus 1 equals negative 2 0 sine P. We can simplify it even further to get negative 2 14 equals negative 2 34. The same peak we finally get cosine of P as equals to 14 divided by 2 34 which we would take the inverse cos to get pe by itself. If we approximate this, we get P is about the angle 23. degrees which rounds to 24 degrees. If we're using call numbers, like our answer has now we can do the same thing for all of our other missing angles. Let's say we want you next. No. If we use a little shortcut here we jump down. Yeah, B squared minus A squared minus C squared divided by negative to ac will be equals to cosine B. Or in other words, using Q be it Q squared minus P squared minus R squared divided by two pr this equals the cosine of Q which now we can use our substitutions nine squared minus six squared minus 13 squared divided by two, multiplied by six and with a negative in front gives us cosine Q. Now to save a little bit of time, I will pres simplify this but we get cosine QE, it equals to 1 24 divided by 1 56. You will then be the inverse cosine of 1 24 divided by 1 56 which gives us about 37.35 degrees. Well, she can round down to 37. Finally, we can do the same for R. We can to assign R S equals to R squared minus P squared minus Q squared divided by negative two PQ, which should give us 13 squared minus six squared minus nine squared divided by negative two, multiplied by six and night. Simplifying this, we will end up with R equals the angle of 119 degrees. Which could have also, if we knew the sum of angles in the triangles. When were 1 80 we could use our other two angles to also get articles 19. We have 1 80 equals 24 plus plus R which should give us R is equals to 1 19 degrees. We now have all of the angles for our answer. We have P equals 24 degrees. Q equals 37 degrees and R equals degrees. If we look at our possible answers, we determine the answer to our problem. His answer C OK. I hope to help you solve the problem. Thanks for watching. Goodbye.

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In Exercises 9–24, solve each triangle. Round lengths to the nearest tenth and angle measures to the nearest degree.a = 5, b = 7, c = 10

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