Law of Cosines - SAT Subject Test in Math II

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Question

A triangle has sides that measure 10, 12, and 16. What is the greatest measure of any of its angles (nearest tenth of a degree)?

Answer

We are seeking the measure of the angle opposite the side of greatest length, 16.

We can use the Law of Cosines, setting , and solving for :

$\cos C = \frac{a^{2}+b^{2}- c^{2}}{2ab}$

$\cos C = \frac{10^{2}+12^{2}- 16^{2}}{2 \cdot 10 \cdot 12}$

$\cos C = \frac{100+144- 256}{240}$

$\cos C = \frac{-12}{240}$

$\cos C =- 0.0500$

$C = \cos ^{-1}(-0.0500) \approx 92.9^{\circ}$

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Question

A triangle has sides that measure 15, 17, and 30. What is the least measure of any of its angles (nearest tenth of a degree)?

Answer

We are seeking the measure of the angle opposite the side of least length, 15.

We can use the Law of Cosines, setting , and solving for :

$\cos C = \frac{a^{2}+b^{2}- c^{2}}{2ab}$

$\cos C = \frac{17^{2}+30^{2}- 15^{2}}{2 \cdot 17 \cdot 30}$

$\cos C = \frac{289+900- 225 }{1,020}$

$\cos C = \frac{964 }{1,020}$

$\cos C \approx 0.9451$

$C \approx \cos ^{-1} 0.9451 \approx 19.1^{\circ}$

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Question

Given : $\Delta ABC$ with $AB = 12, BC = 17, m \angle B= 136 ^{\circ}$ .

Which of the following whole numbers is closest to ?

Answer

Apply the Law of Cosines

$c =\sqrt{ a^{2}+b^{2}- 2ab \cos C}$

setting $a = 12, b= 17, C= 136 ^{\circ}$ and solving for :

$c =\sqrt{ 12^{2}+17^{2}- 2\cdot 12 \cdot 17 \cos 136^{\circ}}$

$c \approx \sqrt{144+289-408 (-0.7193)}$

$c \approx \sqrt{144+289+293.49}$

$c \approx \sqrt{726.49}$

$c \approx 27.0$

Of the five choices, 27 comes closest.

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Question

Given : $\Delta ABC$ with $AB = 12, BC = 16, m \angle B= 56 ^{\circ}$ .

Evaluate to the nearest tenth.

Answer

Apply the Law of Cosines

$c =\sqrt{ a^{2}+b^{2}- 2ab \cos C}$

setting $a = 12, b = 16, C = 56^{\circ}$ and solving for :

$c =\sqrt{ 12^{2}+16^{2}- 2\cdot 12 \cdot 16 \cos 56^{\circ}}$

$c \approx \sqrt{ 144+256- 384 \cdot 0.5592 }$

$c \approx \sqrt{ 144+256- 214.73 }$

$c \approx \sqrt{ 185.27 }$

$c \approx 13.6$

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Question

In $\bigtriangleup ABC$ :

$m \angle A = 105^{\circ }$

Evaluate the length of $\overline{BC}$ to the nearest tenth of a unit.

Answer

The figure referenced is below:

By the Law of Cosines, given the lengths and of two sides of a triangle, and the measure $\gamma$ of their included angle, the length of the third side can be calculated using the formula

$c^{2} = a ^{2} + b^{2} - 2ab \cos \gamma$

Substituting , , , and $\gamma = m \angle A = 105^{\circ }$ , then evaluating:

$(BC)^{2} = 12^{2} + 15 ^{2} - 2 (12)(15)\cos 105^{\circ }$

$\approx 144+225 - 360 (-0.2588 )$

$\approx 369 - (-93.17 )$

$\approx 369 - (-93.17 )$

$\approx 462.17$

Taking the square root of both sides:

$BC\approx \sqrt{462.17} \approx 21.6$

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