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.

b = 5, c = 3, A = 102°

Accepted Answer

Hello everybody. I hope you are doing. All right. Today, today we're going to be looking at this question that states find the missing side length and angles of the triangle and express the length to one decimal place in the angle to the nearest degree. Now, in this question, lower case letters will represent a side and an upper case letter will represent an angle. So the values that they give us are side, Q si is equal to six side, R is equal to two and angle, P is equal to 38 degrees. Now, the answer choice is provided, I'll give us two angles and being Q and R and one side being P. So answer choice A states that P equals 4. Q equals 126 degrees and R equals 16 degrees. As a choice B says that P equals 4.6 Q equals 16 degrees and R equals 126 degrees. C says that P is equal to 9.2 Q is equal to 126 degrees and R is equal to 16 degrees. And as your choice, D says that P equals 9.2 Q equals 16 degrees and R equals 126 degrees. Now, the first thing we'll do here is write some laws that we know with the variables that we are given. Now, specifically, I will be using cosine laws. No, with those laws, I can write three different equations, each one solving for a side. So for equation number one, we'll have side Q squared is equal to side R squared plus side P squared minus two multiplied by side R and side P multiplying the cosine of angle Q. Now, for our second equation, it will have the same general idea as equation number one. But the vari the variables will be moved around a bit. So let's say we're now solving for angle R, this will be angle or side R will have side R squared is equal to side Q squared plus side P squared minus two multiplied by side Q multiplied by side P multiplying the cosine of angle R. And finally, for our third equation, same thing, we'll have the same general idea but we'll move the variables around. So now let's say we're solving for side P squared, this will be equal to side R squared plus side Q squared minus two multiplying by side R and side Q multiplying the cosine of angle P. Now these are laws cosine laws that we can use to use and manipulate to solve for the variables that we have to solve for, for example, to find side ping, we're going to use equation number three cousin equation number three, we have side P squared equal to the rest of the equation. So using equation three, we'll have that side P squared is equal to. And now we can plug in some numbers, we'll have two squared plus six squared minus two, multiplied by two, multiplied by six, multiplied by the cosine of angle P which is given to us and angle P is 38 degrees. So it will be the co sign of 38 degrees. So let's start to simplify. We'll have side P squared is equal to four plus 36 minus 24 multiplied by the cosine of 38 degrees. And if we simplify that we'll have that side P squared is equal to 21.08. And now to solve four side P, we take the square root of both sides. So have that P is equal to the square root of 21.08 which runs off to be around 4.592. And because they asked us to express the length to one decimal place, we'll have that side P is around 4.6. And that will be our answer for side P. And now we can move on and we'll find angle Q. So it's a fine angle cue, we can use one of the formulas and manipulate it for. So ex expression number one has cosine of Q. So if we manipulate it and we can solve for the cosine of U, we'll have that cosine of Q. And we're using equation number one, we'll have cosine of Q is equal to P squared plus R squared minus Q squared. And all of that divided by two multiplied by P multiplied by R. So now we can plug in some numbers and we'll have that cosine of Q is equal to 4.6 squared plus two squared minus six squared. And all of that divided by two, multiplied by 4.6 multiplied by two. Now, if we do some simplification, we'll have that cosine of Q is equal to negative 10.84 divided by 18.4. And now we simplify that we'll have that cosine of Q is equal to negative 0.5891. And now to solve for the angle Q, we can take what we call inverse cosine of the value that we obtained. So the inverse cosine is represented by cosine to a negative one exponent. So we'll have cosine to the negative one exponent of negative 0.5891. And this will be equal to Q. And if we use our calculator to solve for this, we'll have that Q is equal to 126. which rounds off they ask us to around the angle to the nearest degree. So this will round off to be 126 degrees. And we have the answer for angle Q. Now, finally, to find angle R, this will be a much easier process because we have two angles already and we're only missing one. Now, we know that a triangle is equal to 180 degrees. So when you add up all the angles within a triangle, they would add up to 180 degrees. Always. Therefore, angle P plus angle, Q plus angle R is equal to 180 degrees. And we have that 38 degrees plus degrees plus angle R is equal to degrees, meaning that R is equal to 180 degrees minus 38 degrees minus 126 degrees. And if we simplify all that, we're left with angle R equals 16 degrees and that will be our final answer or angle R. And if we look at our answer choices, we see that answer choice a matches with our answer, our answers obtained. So we have side P equals 4.6 units. Angle Q equals 126 degrees and angle R is equal to 16 degrees. So we can see how important the cosine laws are and how useful they can become when applying them to a question like this one where we have to solve for different angles and different sides. So I hope that was helpful and until next time.

In Exercises 9–24, solve each triangle. Round lengths to the nearest tenth and angle measures to the nearest degree. b = 5, c = 3, A = 102°

Key Concepts

Law of Sines

Triangle Sum Theorem

Ambiguous Case of the Law of Sines

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