Solve sqrt{3^2+6^2+2*6*3cos(60)}

$$\sqrt{ { 3 }^{ 2 } + { 6 }^{ 2 } +2 \cdot 6 \cdot 3 \cos ( 60 ) }$$

Evaluate

$$3\sqrt{7}\approx 7.937253933$$

Solution Steps

Calculate 3 to the power of 2 and get 9.

$$\sqrt{9+6^{2}+2\times 3\times 6\cos(60)}$$

Calculate 6 to the power of 2 and get 36.

$$\sqrt{9+36+2\times 3\times 6\cos(60)}$$

Add 9 and 36 to get 45.

$$\sqrt{45+2\times 3\times 6\cos(60)}$$

Multiply 2 and 3 to get 6.

$$\sqrt{45+6\times 6\cos(60)}$$

Multiply 6 and 6 to get 36.

$$\sqrt{45+36\cos(60)}$$

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

$$\sqrt{45+36\times \frac{1}{2}}$$

Multiply 36 and \frac{1}{2} to get 18.

$$\sqrt{45+18}$$

Add 45 and 18 to get 63.

$$\sqrt{63}$$

Factor 63=3^{2}\times 7. Rewrite the square root of the product \sqrt{3^{2}\times 7} as the product of square roots \sqrt{3^{2}}\sqrt{7}. Take the square root of 3^{2}.

$$3\sqrt{7}$$