Solve 19.1-41+3tan60^circ-sqrt{12}-(1/2)-1

$$19.1 - 41 + 3 \tan 60 ^ { \circ } - \sqrt { 12 } - ( \frac { 1 } { 2 } ) - 1$$

Evaluate

$$\sqrt{3}-23.4\approx -21.667949192$$

Solution Steps

Subtract 41 from 19.1 to get -21.9.

$$-21.9+3\tan(60)-\sqrt{12}-\frac{1}{2}-1$$

Get the value of \tan(60) from trigonometric values table.

$$-21.9+3\sqrt{3}-\sqrt{12}-\frac{1}{2}-1$$

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$-21.9+3\sqrt{3}-2\sqrt{3}-\frac{1}{2}-1$$

Combine 3\sqrt{3} and -2\sqrt{3} to get \sqrt{3}.

$$-21.9+\sqrt{3}-\frac{1}{2}-1$$

Subtract \frac{1}{2} from -21.9 to get -\frac{112}{5}.

$$-\frac{112}{5}+\sqrt{3}-1$$

Subtract 1 from -\frac{112}{5} to get -\frac{117}{5}.

$$-\frac{117}{5}+\sqrt{3}$$