In Exercises 1–12, solve each triangle. Round lengths to the near...

Question

In Exercises 1–12, solve each triangle. Round lengths to the nearest tenth and angle measures to the nearest degree. If no triangle exists, state 'no triangle.' If two triangles exist, solve each triangle.

C = 50°, a = 3, c = 1

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 triangle. If they do produce triangles, we need to find the missing side lengths and angles of those triangles and express the lengths and angles in one decimal place. We are given that C is an angle measuring 70 degrees A is four and the side C is two, we have four answer choices, answer choice A and B represent one triangle. Answer choice C represents two triangles and answer choice D represents no triangle. We're gonna use the law of signs. So recall that the law of sign says that the side A divided by the sign of angle A would equal the side B divided by the sign of angle B would equal side C divided by the sign of angle C. Here, we only have A S and CS. So we're gonna just use the A divided by the sign of A is equal to C divided by the sign of C. So plugging in the values we are given A is four. So four divided by the sign of angle A which we don't know. So that's gonna be our unknown for now equals C, which is two divided by the sine of C. So the sine of 70 degrees, this is a proportion. So I'm going to cross multiply and when I do, I have 40 sine of 70 degrees equal to two sine of a, get the sign of A by itself. I'm dividing both sides by two. And this will give me that two sine of 70 degrees will equal the sign of A. So before I go through and use the arc sign, I'm gonna try to find out what value is two sine of 70 degrees because I recall that sine of A has to be less than, or equal to one. So if two multiplied by the sine of 70 degrees is greater than one, we'd have no triangle. So using a calculator, make sure it is in degree mode two multiplied by the sign of degrees is 1.8793. Since that is greater than one, we have no triangle. So that is answer choice. D have a nice day.

In Exercises 1–12, solve each triangle. Round lengths to the nearest tenth and angle measures to the nearest degree. If no triangle exists, state 'no triangle.' If two triangles exist, solve each triangle. C = 50°, a = 3, c = 1

Key Concepts

Law of Sines

Ambiguous Case of the Law of Sines

Triangle Existence Criteria

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