Applying the Law of Cosines - Math

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Question

A triangle has sides of length 12, 17, and 22. Of the measures of the three interior angles, which is the greatest of the three?

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Answer

We can apply the Law of Cosines to find the measure of this angle, which we will call :

The widest angle will be opposite the side of length 22, so we will set:

, ,

$484 = 144 + 289 -408 \cos C$

$51= -408 \cos C$

$\cos C = -$\frac{51}{408}$ = -0.125$

$C = $\cos^{-1}$ 0.125 = $97.2^{\circ }$$

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Question

In $\Delta ABC$ , $m \angle A = $66^{\circ }$$ , , and . To the nearest tenth, what is ?

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Answer

By the Law of Cosines:

$\left ( BC\right $)^{2}$ = \left ( AB\right $)^{2}$ + \left ( AC\right $)^{2}$ - 2 \cdot AB \cdot AC \cdot \cos m \angle A$

or, equivalently,

$BC =$\sqrt{ \left( AB\right $)^{2}$$ + \left( AC\right $)^{2}$ - 2 \cdot AB \cdot AC \cdot \cos m \angle A}$

Substitute:

$BC =$\sqrt{ $26^{2}$$ $+23^{2}$ - 2 \cdot 26 \cdot 23 \cdot \cos 66 ^{\circ }}$

$\approx $\sqrt{ 676 +529 - 1,196 \cdot 0.4067}$$

$\approx $\sqrt{ 718.6}$ \approx 26.8$

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Question

In $\Delta ABC$ , , , and . To the nearest tenth, what is $m \angle A$ ?

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Answer

By the Triangle Inequality, this triangle can exist, since .

By the Law of Cosines:

$\left ( BC\right $)^{2}$ = \left ( AB\right $)^{2}$ + \left ( AC\right $)^{2}$ - 2 \cdot AB \cdot AC \cdot \cos m \angle A$

Substitute the sidelengths and solve for $\cos m \angle A$ :

$1,024 =676 + 529 - 1,196 \cdot \cos m \angle A$

$1,024 =1,205 - 1,196 \cdot \cos m \angle A$

$-181 = - 1,196 \cdot \cos m \angle A$

$\cos m \angle A = 181 \div 1,196 \approx 0.1513$

$m \angle A \approx $\cos^{-1}$ 0.1513 \approx 81.3 ^{\circ }$

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Question

A triangle has sides of length 12, 17, and 22. Of the measures of the three interior angles, which is the greatest of the three?

Tap to reveal answer

Answer

We can apply the Law of Cosines to find the measure of this angle, which we will call :

The widest angle will be opposite the side of length 22, so we will set:

, ,

$484 = 144 + 289 -408 \cos C$

$51= -408 \cos C$

$\cos C = -$\frac{51}{408}$ = -0.125$

$C = $\cos^{-1}$ 0.125 = $97.2^{\circ }$$

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Question

In $\Delta ABC$ , $m \angle A = $66^{\circ }$$ , , and . To the nearest tenth, what is ?

Tap to reveal answer

Answer

By the Law of Cosines:

$\left ( BC\right $)^{2}$ = \left ( AB\right $)^{2}$ + \left ( AC\right $)^{2}$ - 2 \cdot AB \cdot AC \cdot \cos m \angle A$

or, equivalently,

$BC =$\sqrt{ \left( AB\right $)^{2}$$ + \left( AC\right $)^{2}$ - 2 \cdot AB \cdot AC \cdot \cos m \angle A}$

Substitute:

$BC =$\sqrt{ $26^{2}$$ $+23^{2}$ - 2 \cdot 26 \cdot 23 \cdot \cos 66 ^{\circ }}$

$\approx $\sqrt{ 676 +529 - 1,196 \cdot 0.4067}$$

$\approx $\sqrt{ 718.6}$ \approx 26.8$

← Didn't Know|Knew It →

Question

In $\Delta ABC$ , , , and . To the nearest tenth, what is $m \angle A$ ?

Tap to reveal answer

Answer

By the Triangle Inequality, this triangle can exist, since .

By the Law of Cosines:

$\left ( BC\right $)^{2}$ = \left ( AB\right $)^{2}$ + \left ( AC\right $)^{2}$ - 2 \cdot AB \cdot AC \cdot \cos m \angle A$

Substitute the sidelengths and solve for $\cos m \angle A$ :

$1,024 =676 + 529 - 1,196 \cdot \cos m \angle A$

$1,024 =1,205 - 1,196 \cdot \cos m \angle A$

$-181 = - 1,196 \cdot \cos m \angle A$

$\cos m \angle A = 181 \div 1,196 \approx 0.1513$

$m \angle A \approx $\cos^{-1}$ 0.1513 \approx 81.3 ^{\circ }$

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Question

A triangle has sides of length 12, 17, and 22. Of the measures of the three interior angles, which is the greatest of the three?

Tap to reveal answer

Answer

We can apply the Law of Cosines to find the measure of this angle, which we will call :

The widest angle will be opposite the side of length 22, so we will set:

, ,

$484 = 144 + 289 -408 \cos C$

$51= -408 \cos C$

$\cos C = -$\frac{51}{408}$ = -0.125$

$C = $\cos^{-1}$ 0.125 = $97.2^{\circ }$$

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Question

In $\Delta ABC$ , $m \angle A = $66^{\circ }$$ , , and . To the nearest tenth, what is ?

Tap to reveal answer

Answer

By the Law of Cosines:

$\left ( BC\right $)^{2}$ = \left ( AB\right $)^{2}$ + \left ( AC\right $)^{2}$ - 2 \cdot AB \cdot AC \cdot \cos m \angle A$

or, equivalently,

$BC =$\sqrt{ \left( AB\right $)^{2}$$ + \left( AC\right $)^{2}$ - 2 \cdot AB \cdot AC \cdot \cos m \angle A}$

Substitute:

$BC =$\sqrt{ $26^{2}$$ $+23^{2}$ - 2 \cdot 26 \cdot 23 \cdot \cos 66 ^{\circ }}$

$\approx $\sqrt{ 676 +529 - 1,196 \cdot 0.4067}$$

$\approx $\sqrt{ 718.6}$ \approx 26.8$

← Didn't Know|Knew It →

Question

In $\Delta ABC$ , , , and . To the nearest tenth, what is $m \angle A$ ?

Tap to reveal answer

Answer

By the Triangle Inequality, this triangle can exist, since .

By the Law of Cosines:

$\left ( BC\right $)^{2}$ = \left ( AB\right $)^{2}$ + \left ( AC\right $)^{2}$ - 2 \cdot AB \cdot AC \cdot \cos m \angle A$

Substitute the sidelengths and solve for $\cos m \angle A$ :

$1,024 =676 + 529 - 1,196 \cdot \cos m \angle A$

$1,024 =1,205 - 1,196 \cdot \cos m \angle A$

$-181 = - 1,196 \cdot \cos m \angle A$

$\cos m \angle A = 181 \div 1,196 \approx 0.1513$

$m \angle A \approx $\cos^{-1}$ 0.1513 \approx 81.3 ^{\circ }$

← Didn't Know|Knew It →

Question

A triangle has sides of length 12, 17, and 22. Of the measures of the three interior angles, which is the greatest of the three?

Tap to reveal answer

Answer

We can apply the Law of Cosines to find the measure of this angle, which we will call :

The widest angle will be opposite the side of length 22, so we will set:

, ,

$484 = 144 + 289 -408 \cos C$

$51= -408 \cos C$

$\cos C = -$\frac{51}{408}$ = -0.125$

$C = $\cos^{-1}$ 0.125 = $97.2^{\circ }$$

← Didn't Know|Knew It →

Question

In $\Delta ABC$ , $m \angle A = $66^{\circ }$$ , , and . To the nearest tenth, what is ?

Tap to reveal answer

Answer

By the Law of Cosines:

$\left ( BC\right $)^{2}$ = \left ( AB\right $)^{2}$ + \left ( AC\right $)^{2}$ - 2 \cdot AB \cdot AC \cdot \cos m \angle A$

or, equivalently,

$BC =$\sqrt{ \left( AB\right $)^{2}$$ + \left( AC\right $)^{2}$ - 2 \cdot AB \cdot AC \cdot \cos m \angle A}$

Substitute:

$BC =$\sqrt{ $26^{2}$$ $+23^{2}$ - 2 \cdot 26 \cdot 23 \cdot \cos 66 ^{\circ }}$

$\approx $\sqrt{ 676 +529 - 1,196 \cdot 0.4067}$$

$\approx $\sqrt{ 718.6}$ \approx 26.8$

← Didn't Know|Knew It →

Question

In $\Delta ABC$ , , , and . To the nearest tenth, what is $m \angle A$ ?

Tap to reveal answer

Answer

By the Triangle Inequality, this triangle can exist, since .

By the Law of Cosines:

$\left ( BC\right $)^{2}$ = \left ( AB\right $)^{2}$ + \left ( AC\right $)^{2}$ - 2 \cdot AB \cdot AC \cdot \cos m \angle A$

Substitute the sidelengths and solve for $\cos m \angle A$ :

$1,024 =676 + 529 - 1,196 \cdot \cos m \angle A$

$1,024 =1,205 - 1,196 \cdot \cos m \angle A$

$-181 = - 1,196 \cdot \cos m \angle A$

$\cos m \angle A = 181 \div 1,196 \approx 0.1513$

$m \angle A \approx $\cos^{-1}$ 0.1513 \approx 81.3 ^{\circ }$

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