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Problem 62

Determine whether each statement makes sense or does not make sense, and explain your reasoning. If I know the measures of all three angles of an oblique triangle, neither the Law of sines nor the Law of Cosines can be used to find the length of a side.

Problem 64

Determine whether each statement makes sense or does not make sense, and explain your reasoning. Solving an SSS triangle, I do not have to be concerned about the ambiguous case when using the Law of sines.

Problem 64

Convert each polar equation to a rectangular equation. Then use a rectangular coordinate system to graph the rectangular equation. $$ r \cos \theta=7 $$

Problem 66

Explaining the Concepts. Why is SSA called the ambiguous case?

Problem 71

In Exercises 71–74, determine whether each statement makes sense or does not make sense, and explain your reasoning. I began using the Law of Sines to solve an oblique triangle in which the measures of two sides and the angle between them were known.

Problem 74

In Exercises 71–74, determine whether each statement makes sense or does not make sense, and explain your reasoning. Under certain conditions, a fire can be located by superimposing a triangle onto the situation and applying the Law of Sines.

Problem 75

If you are given two sides of a triangle and their included angle, you can find the triangle’s area. Can the Law of Sines be used to solve the triangle with this given information? Explain your answer.

Problem 76

Find the smallest interval for \(\theta\) starting with \(\theta \min =0\) so that your graphing utility graphs the given polar equation exactly once without retracing any portion of it. $$r=4 \sin \theta$$

Problem 77

In Exercises 77–78, round answers to the nearest pound. a. Find the magnitude of the force required to keep a 3500 -pound car from sliding down a hill inclined at \(5.5^{\circ}\) from the horizontal. b. Find the magnitude of the force of the car against the hill.

Problem 81

In Exercises \(81-86,\) solve equation in the complex number system. Express solutions in polar and rectangular form. $$ x^{6}-1=0 $$

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