Law of Cosines - Math

Card 0 of 12

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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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 }$$

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

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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$

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

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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 }$

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 see back →

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 }$$

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

Tap to see back →

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$

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

Tap to see back →

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 }$

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 see back →

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 }$$

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

Tap to see back →

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$

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

Tap to see back →

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 }$

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 see back →

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 }$$

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

Tap to see back →

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$

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

Tap to see back →

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 }$