# Trigonometrysolving for postive value of x when cos (60)

#### gotah

##### New member
Hello guys!

I'm new to this forum, so I apologize if this Q is in the wrong thread -it's a mixture of trigonometry and algebra. I've tried to solve it but i don't get the correct answer. I am given a triangle, whose sides i had to calculate, which i have done, and the results are correct.
so the triangle OPA has sides: OP= sqr(x^2 + 100) and PA = sqr(x^2 -16x +80), and AO=10
Thereafter i have to show that cos OPA = cos(60°)=((x^2)-8x+40) / sqr(((x^2)-16x+80)((x^2)+100)), which i have also done.

After a few other Q I am asked to find the positive value of x such that the angle OPA = 60degrees, which is the problem i cannot solve. I have tried to solve for x with the above equation but it just becomes too complicated.

I have also tried to just solve for x by saying sin(60degrees)= 10 / sqr(x^2 +100) , where i get x=5.77 but the result should be 5.63.

Is there any other way to find the positive value than having to solve for x in the complicated equation above?

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
find the positive value of x such that the angle OPA = 60degrees
Do you need to solve this analytically or numerically?

I have also tried to just solve for x by saying sin(60degrees)= 10 / sqr(x^2 +100) , where i get x=5.77 but the result should be 5.63.
There is no reason to believe that angle OAP = 90 degrees.

#### gotah

##### New member
i am not allowed to use the calculator, so it would be both? When i use the cosine rule, I get stuck after expanding the bracket.

(using the cosine rule i've said
10^2 = [sqr(x^2 -16x+80)]^2 + [sqr(x^2 +100)]^2 - 2*sqr(x^2 -16x+80)* sqr(x^2 +100) * cos (60))

(x^2-16x+80 +x^2+100-100) / (2*sqr(x^2 -16x+80)* sqr(x^2 +100)= cos(60) =0.5

(x^2-16x+80 +x^2+100-100) = sqr(x^2 -16x+80)* sqr(x^2 +100)

2x^2-16x+80 = sqr(x^2 -16x+80)* sqr(x^2 +100)

(2x^2-16x+80)^2 = (x^2 -16x+80)* (x^2 +100)

4x^4-64x^3+576x^2-2560x+6400=x^4-16x^3+180x^2-1600x+8000

3x^4-48x^3+396x^2-960x-1600 =0 (how to solve this for x? I dunno how to factorize it)

Last edited:

#### skeeter

##### Well-known member
MHB Math Helper
so the triangle OPA has sides: OP= sqr(x^2 + 100) and PA = sqr(x^2 -16x +80), and AO=10
Thereafter i have to show that cos OPA = cos(60°)=((x^2)-8x+40) / sqr(((x^2)-16x+80)((x^2)+100)), which i have also done.

After a few other Q I am asked to find the positive value of x such that the angle OPA = 60degrees, which is the problem i cannot solve. I have tried to solve for x with the above equation but it just becomes too complicated.
10^2 = (x^2+100) + (x^2-16x+80) - 2sqrt{(x^2+100)(x^2-16x+80)} cos(60)

0 = (x^2+100) + (x^2-16x+80) - sqrt{(x^2+100)(x^2-16x+80)} - 100

solved by calculator, x = 5.6341956 ...

#### gotah

##### New member
thank you for the effort, but i'm not allowed to use the calculator:-(

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
i am not allowed to use the calculator
The exact solution is $x = 4-\sqrt{3}+\sqrt{7 \left(\frac{8}{\sqrt{3}}-3\right)}$. Suppose that you get this answer. Are you really expected to make sure that this is 5.63 without a calculator?

Perhaps it would help to know the background of this problem. You said that you had to calculate the sides. What is x and what is the whole problem statement?

#### gotah

##### New member
ok. Two lines are drawn in a diagram. Point O and A are fixed and have coordinates (0,0), (8,6) respectively. The Angle OPA varies as the Point P(x,10) move along the horizontal line y=10.

first i have to show that: 1. AP = sqr(x^2 -16x + 80)
2.then I have to find a similar expression for OP in terms of x,
3. hence show that the angle cos(OPA) = (x^2-8x +40) / sqr(x2-16x+80)(X^2 + 100)

4.Then it asks me to find the angle OPA when x= 8
5.And then Find the positive value of x such that OPA = 60degrees.
6. Let function f(x) be cos(OPA) = (x^2-8x +40) / sqr(x2-16x+80)(X^2 + 100) , 0_<x<_15
and consider the f(x)=1. i.Explain, in terms of the position of points O,A and P why this has a solution.

Maybe i should just use the calculator, but my book states when I'm allowed to do so, and it didn't with this Q. ( i also don't understand no. 6 by the way. is it because if i draw a horizontal line Y=1, it will intersect with the triangle? )

Last edited:

#### gotah

##### New member
i found the solution for this on another forum (cannot post the link since i don't have 10posts, but it's in mymathforum. com), and i want to thank everyone who tried to help me.although, i still have to study the solution