Precalculus Examples

$b^{2} = 32^{2} + {60.5}^{2} - 2 \cdot 32 \cdot 60.5 \cos (19 °)$

Step 1

Multiply $- 2$ by $32$ .

$b^{2} = 32^{2} + {60.5}^{2} - 64 \cdot (60.5 \cos (19 °))$

Step 2

Evaluate $\cos (19 °)$ .

$b^{2} = 32^{2} + {60.5}^{2} - 64 \cdot (60.5 \cdot 0.94551857)$

Step 3

Multiply $60.5$ by $0.94551857$ .

$b^{2} = 32^{2} + {60.5}^{2} - 64 \cdot 57.20387382$

Step 4

Multiply $- 64$ by $57.20387382$ .

$b^{2} = 32^{2} + {60.5}^{2} - 3661.04792472$

Step 5

Subtract $3661.04792472$ from $32^{2}$ .

$b^{2} = {60.5}^{2} - 2637.04792472$

Step 6

Subtract $2637.04792472$ from ${60.5}^{2}$ .

$b^{2} = 1023.20207527$

Step 7

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$b = \pm \sqrt{1023.20207527}$

Step 8

The complete solution is the result of both the positive and negative portions of the solution.

Tap for more steps...

Step 8.1

First, use the positive value of the $\pm$ to find the first solution.

$b = \sqrt{1023.20207527}$

Step 8.2

Next, use the negative value of the $\pm$ to find the second solution.

$b = - \sqrt{1023.20207527}$

Step 8.3

The complete solution is the result of both the positive and negative portions of the solution.

$b = \sqrt{1023.20207527}, - \sqrt{1023.20207527}$

Step 9

The result can be shown in multiple forms.

Exact Form:

$b = \sqrt{1023.20207527}, - \sqrt{1023.20207527}$

Decimal Form:

$b = 31.98752999 \dots, - 31.98752999 \dots$