8. Find the value of x. If your answer is not an integer, leave it in simplest radical form. The diagram is not drawn to scale.
![8 Find the value of x If your answer is not an integer leave it in simplest radical form The diagram is not drawn to scale class=](https://us-static.z-dn.net/files/d57/a625222815578730ee841defe28946cf.png)
Here in the figure, a right angled triangle given.
One angle of other two angles given [tex] 30^o [/tex].
The hypotenuse of an right angled triangle is always opposite to the right angle. So here, the side of length 18 is hypotenuse..
The side of length 9 is opposite to the ange given [tex] 30^o [/tex].
So here two sides given and we need to find the third side.
We can get the third side by using either a trigonometric function or by using Pythagoras theorem.
Let's use Pythagoras theorem here.
We know the theorem is if a and b are other two sides and c is the hypotenuse then, [tex] c^2 = a^2+b^2 [/tex]
Here, c = 18, and we can take b = 9 and a is x here. By substituting the values in the formula we will get,
[tex] 18^2 = x^2 + 9^2 [/tex]
[tex] 324 = x^2+81 [/tex]
Now to get x, we will move 81 to the left side by subtracting it from both sides. We will get,
[tex] 324 - 81 = x^2+81-81 [/tex]
[tex] 324-81 = x^2 [/tex]
[tex] 243 = x^2 [/tex]
[tex] x^2 = 243 [/tex]
We acn get x by taking square root to both sides. We will get,
[tex] \sqrt{x^2} = \sqrt{243} [/tex]
[tex] x = \sqrt{243} [/tex]
We have to simplify 243 now. We can write 243 as a multiplication of 81 and 3.
[tex] x = \sqrt{(81)(3)} [/tex]
[tex] x = (\sqrt{81})(\sqrt{3}) [/tex]
[tex] x = 9\sqrt{3} [/tex]
So we have got the required answer here. The side x = [tex] 9\sqrt{3} [/tex]