Solve 40/sqrt{3}-40/tan45^circ

$$40 / \sqrt { 3 } - 40 / \tan 45 ^ { \circ }$$

Evaluate

$$\frac{40\left(\sqrt{3}-3\right)}{3}\approx -16.905989232$$

Short Solution Steps

Rationalize the denominator of \frac{40}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\frac{40\sqrt{3}}{\left(\sqrt{3}\right)^{2}}-\frac{40}{\tan(45)}$$

The square of \sqrt{3} is 3.

$$\frac{40\sqrt{3}}{3}-\frac{40}{\tan(45)}$$

Get the value of \tan(45) from trigonometric values table.

$$\frac{40\sqrt{3}}{3}-\frac{40}{1}$$

Anything divided by one gives itself.

$$\frac{40\sqrt{3}}{3}-40$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 40 times \frac{3}{3}.

$$\frac{40\sqrt{3}}{3}-\frac{40\times 3}{3}$$

Since \frac{40\sqrt{3}}{3} and \frac{40\times 3}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{40\sqrt{3}-40\times 3}{3}$$

Do the multiplications in 40\sqrt{3}-40\times 3.

$$\frac{40\sqrt{3}-120}{3}$$