Precalculus Law of Sines and Law of Cosines Quiz: 10 Questions with Explanations

A Law of SinesB Pythagorean theoremC Law of CosinesD Heron’s formula

SAS has no angle-side pair, so the Law of Cosines is the starting point.

A \(10\sqrt2\)B \(5\sqrt2\)C \(10\)D \(20\)

b = 10 sin45 / sin30 = 10(sqrt2/2)/(1/2) = 10 sqrt 2.

A 1B 2C 0D 3

Sine cannot exceed 1, so no triangle exists.

A The SAS caseB The ambiguous case (SSA)C A right triangle onlyD The SSS case

The side opposite the given angle can reach the base in two places.

A 24B \(24\sqrt3\)C 48D 12

Area = 1/2 (6)(8) sin 30 = 24 times 1/2 = 12.

A 42B 168C 210D 84

s = 21 and the area is sqrt(21 times 8 times 7 times 6) = 84.

A \(\sqrt{5^{2}+7^{2}-2(5)(7)\cos40^{\circ}}\)B \(\sqrt{5^{2}+7^{2}+2(5)(7)\cos40^{\circ}}\)C \(\frac{5\sin40^{\circ}}{7}\)D \(5+7-\cos40^{\circ}\)

The Law of Cosines subtracts 2ab cos C.

A Law of CosinesB Law of SinesC Right-triangle sineD None; it is impossible

Two sides and the included angle are known, so use the Law of Cosines.

A The shortest sideB The longest sideC The middle sideD It depends on the triangle

The largest angle always faces the longest side.

A \(c^{2}=a^{2}-b^{2}\)B \(c^{2}=a^{2}+b^{2}\)C \(c=a+b\)D \(c^{2}=2ab\)

cos 90 = 0, so the extra term vanishes, leaving the Pythagorean theorem.