Q:

# angelique draws triangle ghk. if angle g=30o, g=3, ang k=4, what is the approximate length of hA. 1.2 or 5.7B. 2.2 or 4.7C. 4.7D. 5.7

Accepted Solution

A:
To solve this problem we will use the cosine rule. Formula is:
$$x^{2} = y^{2} + z^{2} -2*y*z*cos \alpha$$
On left side we have side that we want to find length of. On right side we have other two sides and angle opposite to searched side.

We are given:
angle g=30Β°
g = 3
k = 4

In case of our formula we know x and y, but we do not know z. Now we have:
$$3^{2} = 4^{2} + z^{2} -2*4*z*cos 30$$
$$9 = 16 + z^{2} -2*4*z* \frac{ \sqrt{3} }{2} \\ 9=16+z^{2} -4\sqrt{3} z \\ z^{2}-4\sqrt{3} z+7=0$$

Now we solve this for z:
$$c_{1} = \frac{-b+ \sqrt{ b^{2}-4ac} }{2a} \\ c_{1} = \frac{4 \sqrt{3}+ \sqrt{48-28} }{2} \\ c_{1} = \frac{4 \sqrt{3}+ \sqrt{20} }{2} \\ c_{1} = \frac{4 \sqrt{3}+ 2\sqrt{5} }{2} \\ c_{1} =2 \sqrt{3} + \sqrt{5} =5.7$$

$$c_{2} = \frac{-b- \sqrt{ b^{2}-4ac} }{2a} \\ c_{2} = \frac{4 \sqrt{3}- \sqrt{48-28} }{2} \\ c_{2} = \frac{4 \sqrt{3}- \sqrt{20} }{2} \\ c_{2} = \frac{4 \sqrt{3}- 2\sqrt{5} }{2} \\ c_{2} =2 \sqrt{3} - \sqrt{5} =1.2$$

Our solution is A.