Trigonometry Triangle Non-Routine Problem

[mathangel]

I found this question in a book which is catergorised under non-routine problem.
The question is:

A triangle ABC, given its area is square root of 3, angle BAC is 60 degree, AB=x, BC=5cm and CA=y. Hence, find the value of x+y.

I am a Malaysian High School student, so maybe the question will look easier than the question from futher studies.

I have done this question many times with many different solutions in papers. I can only type out my solutions that I remembered. In the following solutions, I may use * to represent multiply operation and / to represent divide operation.

Solutions:
( This is to prove that xy=4)
By using the formula of area of triangle,
1/2*ab*sinC=square root of 3

By taking angle BAC=60 degree as C, AB=x cm as a and CA=y cm as b,
Hence, xy=4

By taking BC=5cm as a, AB=y cm as b and leaving C as an unknown angle,
Hence sinC= (2*square root of 3)/5y -(1)

By using sine rule,
5/sin60' = x/sinC -(2)

Subsitute equation (1) into equation (2) and replace sin60' with (square root of 3)/2:
10/(square root of 3) = 5xy/(2*square root of 3)

By comparing both sides of the above equation,
2*10=5xy
Hence, xy=4 [xy=4 is proven]

When I have proven that xy=4, I have stucked in this step. With xy=4, I can't proceed to any further solutions because any one of the variable will be inverted. If it is x/y=4, maybe I can form a quadratic equation to solve the other variable then I can get the value of x+y.
Maybe there are ways to continue that I have forgotten.
By here, I will provide the final answer provided in the book.
The value of x+y = (square root of 37).

Thank you.

 

[pka]

Now use the law of cosines.
\(\displaystyle 25 = x^2 + y^2 - 2xy\cos (60^ \circ )\)

 

[mathangel]

Now use the law of cosines.
\(\displaystyle 25 = x^2 + y^2 - 2xy\cos (60^ \circ )\)

Thanks. This is what the others told me to do.

 

[mathangel]

Good Day everybody!
The point that lead to the final answer is pointed out by my teacher.

After proving xy=4,
By using cosine rule,

a^2=b^2+c^2-2bc cos A

Taking a=5cm, b=x cm, c=y cm and A=60',
Hence, 5^2=x^2+y^2-2xy cos 60
By replacing cos 60=0.5,
Hence, 25=x^2+y^2-xy

By adding 3xy to both sides (+3xy),
Hence, 25+3xy=x^2+y^2-xy+3xy
25+3xy=x^2+y^2+2xy

Rearrange the right hand side of the equation,
Hence, 25+3xy=x^2+2xy+y^2.

By replacing xy=4 at the left hand side of the equation,
25+12=x^2+2xy+y^2

Right hand side of the equation x^2+2xy+y^2 can be simplified to (x+y)^2,
Hence, 37=(x+y)^2

By shifting the square to the left hand side,
Then the answer is obtained,
square root of 37=x+y