Solve sin60^circcos30^circ+tan45^circ

$$\sin 60 ^ { \circ } \cos 30 ^ { \circ } + \tan 45 ^ { \circ }$$

Evaluate

$$\frac{7}{4}=1.75$$

Solution Steps

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

$$\frac{\sqrt{3}}{2}\cos(30)+\tan(45)$$

Get the value of \cos(30) from trigonometric values table.

$$\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}+\tan(45)$$

Multiply \frac{\sqrt{3}}{2} and \frac{\sqrt{3}}{2} to get \left(\frac{\sqrt{3}}{2}\right)^{2}.

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

To raise \frac{\sqrt{3}}{2} to a power, raise both numerator and denominator to the power and then divide.

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

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

$$\frac{\left(\sqrt{3}\right)^{2}}{2^{2}}+1$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{2^{2}}{2^{2}}.

$$\frac{\left(\sqrt{3}\right)^{2}}{2^{2}}+\frac{2^{2}}{2^{2}}$$

Since \frac{\left(\sqrt{3}\right)^{2}}{2^{2}} and \frac{2^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(\sqrt{3}\right)^{2}+2^{2}}{2^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{3}{2^{2}}+1$$

Calculate 2 to the power of 2 and get 4.

$$\frac{3}{4}+1$$

Add \frac{3}{4} and 1 to get \frac{7}{4}.

$$\frac{7}{4}$$