In Exercises 43–44, use the given measurements to solve the follo...

Question

In Exercises 43–44, use the given measurements to solve the following triangle. Round lengths of sides to the nearest tenth and angle measures to the nearest degree.

a = 400, b = 300

Accepted Answer

Find the missing side links. And angles of the triangle express the length and angles in one decimal place where we have Q equals 4 50 P equals 2 50. And we have a triangle here size PQ and R and angles PQ and R. And we also have four answers all with angles PQ and R and lowercase R as a side length. Now to solve this, we can make use of the law of science. La Signs tells us that we can create ratios to find missing sides and angles of a triangle. For example, we'll say side P has a corresponding angle of sine uppercase P which is the same for the others. Side Q is divided by sign of big Q and the same for R. Now we know what P and Q are. We know Q is 4 50 P is 2 50. We will say 2 50 divided by sign of angle P equals 4 50 divided by a sign of angle Q. We also notice angle P. This measure is beta, angle Q's measure is two beta. And we don't have our, let's use our first two ratios to solve for beta. We have 2 50 divided by sine beta. S equals to 4 50 divided by sine two beta. We can cry you get 2 50 sine two beta S equals to 4 50 sign beta. Now we noticed we have a double angle for sun. This is one of our known identities. What we have sun of two X. It's equals to two sine X cosine X. This can then be rewritten as 500 sine beta cosine beta. This equals to 4 50 be, let's now go ahead and move over our 4 50 sine beta. By subtracting we get 500 sin beta cosine beta minus 4 50 sin beta is equals to zero. We can then factor out assigned beta. We can also factor out a number from our 500 our 4 50. The common factor between those two, it's 50 giving us 10 cosine beta minus nine inside parentheses equals zero. From here. We can split this up and solve for beta. We get 50 sine beta equals zero and 10 cosine beta minus nine equals zero. No, our first equation here beta equals zero. Now beta is in grease which should give us a zero degree angle. So we will actually throw this one out. Let's look at our other one. We would get 10 0 beta equals nine. Then we get cosine beta equals nine divided by 10. Now we can take the inverse cosine of both sides giving us beta equals cosine in verse of nine divided by 10, which approximates to be about 25.8 degrees. So now we have our angle beta, we also know angle Q is equals to two beta. This would be two multiplied by 25. which gives us 51.6 degrees. Now finally, we know the sum of angles in the triangle equals 180 degrees. We can say 1 80 equals P plus Q plus R. Now we know P which is 25.8 and we know Q which is 51.6. We can now solve for art, solving for angle R would give us 102.6 degrees. Now finally, we can make use of live signs one more time to find our missing side. R. Let's see, we use 2 50 divided by sine beta, which we now know is 25.8 degrees. S equals to side R divided by sine of 102.6 degrees can cross the multi again, which will eventually give us R is equals to 2 sine 102.6 degrees divided by the sine of 25.8 degrees. Finally, solving for R gives us a side length of 560.6 units. We now have everything we need for our answer. You have angle P, it's 25.8 degrees. Angle Q is 51.6 degrees angle R is 102.6 degrees and side R is 60.6 units. This would be the answer to our problem, which means the answer to our problem turns out to be answer A ok. I hope to help you solve the problem. Thank you for watching. Goodbye.

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In Exercises 43–44, use the given measurements to solve the following triangle. Round lengths of sides to the nearest tenth and angle measures to the nearest degree. a = 400, b = 300