SOLUTION: Triangle. ABC. Area. 100sq metres. Has angles 50 60 & 70. Degrees. Find the. Lengths. Of sides.

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Question 1032587: Triangle. ABC. Area. 100sq metres. Has angles 50 60 & 70. Degrees. Find the. Lengths. Of sides.



Found 2 solutions by Alan3354, ikleyn:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Triangle. ABC. Area. 100sq metres. Has angles 50 60 & 70. Degrees. Find the. Lengths. Of sides.
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To solve this, assign a length to 1 of the sides, then find the other 2 using the Law of Sines.
Then find the area of that triangle.
Then use the fact that the area is a function of the square of side length, and find the sides for area = 100 sq meters.

Answer by ikleyn(52832) About Me  (Show Source):
You can put this solution on YOUR website!
.
A triangle ABC has the area of 100 m%5E2. It has angles of 50°, 60° and 70°.
Find the lengths of the sides of the triangle.
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Let "a", "b" and "c" be the sides of the triangle in a way that the side "a" is opposite to the angle of 60°, 
the side "b" is opposite to the angle 50° and the side "c" is opposite to the angle 70°.

We can write these three equality for the area of the triangle:

%281%2F2%29%2Aa%2Ab%2Asin%2870%5Eo%29 = 100,    (1)

%281%2F2%29%2Ab%2Ac%2Asin%2860%5Eo%29 = 100,    (2)

%281%2F2%29%2Aa%2Ac%2Asin%2850%5Eo%29 = 100.    (3)

Now, multiply all three equality (1), (2) and (3) (both sides). You will get

%281%2F8%29%2Aa%5E2%2Ab%5E2%2Ac%5E2%2Asin%2850%5Eo%29%2Asin%2860%5Eo%29%2Asin%2870%5Eo%29 = 100%5E3.

It gives 

a%5E2%2Ab%5E2%2Ac%5E2 = %288%2A100%5E3%29%2F%28sin%2850%5Eo%29%2Asin%2860%5Eo%29%2Asin%2870%5Eo%29%29,

or, after taking the square root from both sides,

a%2Ab%2Ac = %282%2Asqrt%282%29%2A1000%29%2Fsqrt%28sin%2850%5Eo%29%2Asin%2860%5Eo%29%2Asin%2870%5Eo%29%29.    (4)

Next, let us rewrite the equalities (1) - (3) in the form

a%2Ab = 200%2Fsin%2870%5Eo%29,          (1')

b%2Ac = 200%2Fsin%2860%5Eo%29,          (2')

a%2Ac = 200%2Fsin%2850%5Eo%29.          (3')

We are just at the finish line. Divide (4) by (1'). You will get

c =  = 10%2Asqrt%282%29%2Asqrt%28sin%2870%5Eo%29%2F%28sin%2850%5Eo%29%2Asin%2860%5Eo%29%29%29.      (5)

Divide (4) by (2'). You will get

a =  = 10%2Asqrt%282%29%2Asqrt%28sin%2860%5Eo%29%2F%28sin%2850%5Eo%29%2Asin%2870%5Eo%29%29%29.      (6)

Finally, divide (4) by (3'). You will get

b =  = 10%2Asqrt%282%29%2Asqrt%28sin%2850%5Eo%29%2F%28sin%2860%5Eo%29%2Asin%2870%5Eo%29%29%29.      (7)

Formulas (5), (6) and (7) are the solution of the problem. They allow calculate the sides.
If you want to get numerical values, use  sin(50°) = 0.766044,  sin(60°) = 0.866025  and  sin(70°) = 0.939693. From (5), (6) and (7) you will get

a = 15.5117,  b = 13.7209  and  c = 16.8312  (approximately).