In Exercises 17–32, two sides and an angle (SSA) of a triangle...

Question

In Exercises 17–32, two sides and an angle (SSA) of a triangle are given. Determine whether the given measurements produce one triangle, two triangles, or no triangle at all. Solve each triangle that results. Round to the nearest tenth and the nearest degree for sides and angles, respectively.
a = 30, b = 40, A = 20°

a = 30, b = 40, A = 20°

Accepted Answer

Hello, everyone. We are asked to identify whether the following measurements of two sides in an angle can produce one triangle, two triangles or no triangles. We are going to find the missing side lengths and angles of the triangle. We express the lengths and angles in one decimal place. We are given side A measures side B measures 50 an angle A equals 25 degrees. Our answer choices A and B both describe one triangle with differing values for the angles and sides. Inter choice C describes two triangles and answer choice. D says that we have no triangle to begin since we do not know that this is a right triangle or we don't know anything really else about it. We're gonna use the law of signs. Recall that the law of signs says that we can say that A as a side divided by the sign of angle A will equal B divided by the sign of angle B which will equal C divided by the sign of angle C. So to begin with, I notice we have A S and BS. So I'm gonna start with doing the sign of angle A. So I have the side A just 35 divided by the sine of angle A which is 25 degrees equals the side B which is 50 divided by the sign of angle B which is our unknown. We don't know what the measure of B is. Cross multiplying I get multiplied by the sine of 25 degrees equals 35 multiplied by the sign of angle B. I'm gonna divide both sides by 35 to get the sign of angle B by itself. And using a calculator, I'm going to do 50 multiplied by the sine of 25 degrees, all of which is divided by 35. And when I do so I get the decimal 0. equals the sign of angle B. Since this is less than one and greater than negative one, I know that we can do this and we will actually get a value. So it is not answer choice D also because this is positive. I know there's the possibility of two triangles, one in quadrant one and one in quadrant two. So I'm gonna write that down here since sign of B is positive, we could have a triangle in quadrant one and quadrant two. So let's find out what the angle for B is. So doing the inverse sign of 0.6037 I get that angle B has a measure of 37. degrees. So now knowing that I have two triangles. I'm gonna start with quadrant one. I now know in quadrant one, the angle B is 37.1 degrees. So I can find C by using the law, I'm sorry, the interior angle theorem that says that all triangles, the interior angles add up to 100 and 80 degrees. So I can do 180 degrees minus my value of B 37.1 degrees minus the value of angle A which is 25 degrees. And that will tell me that the angle at sea is 117.9 degrees. So I'm gonna write that up here just so all my information is in one spot. And now that I know two of the angles, we were given two of the sides, I do need to find the third side. So I'm gonna find side C and I'm gonna do that by using side A divided by the sign of angle A equals side B, I'm sorry, side C divided by the sign of angle C. So this will have 35 divided by the sine of 25 degrees equals C divided by the sine of 117. degrees. Cross multiplying I have C multiplied by the sine of 25 degrees equals 35 multiplied by the sine of 117.9 degrees. I'm dividing both sides by the sine of 25 degrees to get sea by itself. And then I will use a calculator, make sure your calculator is in degree mode or you're not gonna get answers that make sense. So when I put this all on a calculator, I get that C is 73 0.2. So that's our length for side C 73.2. So this is one triangle. But as I mentioned, sine is positive and sine can be positive in both quadrant one and quadrant two. So we need to think about another triangle and see if it's possible in quadrant two. If it is, it'll be our first B subtracted from 100 80 degrees. So 180 degrees minus 37. degrees will give us B for quadrant two, which is 142.9 degrees. We'll call that B sub two for now. So then we'll use the same method we did to find C four to find C sub two. We'll do 100 80 degrees minus our value of B. So 142.9 degrees minus the value of angle A which is 25 degrees in subtracting all of those, we are left with C sub two is 12. degrees. So I'm gonna put that right here just so it's again, everything's together and then we need to find side C in the second quadrant. So I'll set it up the same way. So I'm gonna do a divided by the sign of A equals C divided by the sign of C. So 35 divided by the sine of degrees equals C divided by the sine of 12.1 degrees cross multiply and I get C multiplied by the sign of degrees equals 35 multiplied by the sign of 12.1 degrees dividing both sides by the sine of 25 degrees. That'll cancel it out, leave sea by itself. And when I use a calculator to find this value, I find that C is 17.4. So side C or C sub two is 17.4. So this should be the measure for our second triangle. So let's see if everything matches up with answer choice C which is for two triangles. Let's see for two triangles. B one, yep matches up as 37.1. C sub 1, 117.9. Correct side C sub 1 73. second triangle beasts of 2, 142.9 degrees. C sub 2, 12.1 degrees. Side C sub 2, 17.4. So it is answer choice C there are two triangles have a nice day.

Key Concepts

Law of Sines

Ambiguous Case of SSA Triangles

Triangle Solution and Rounding

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