How to find side length of triangle given the three angles and area

Question

Find side length of triangle.

Angle A=50°

Angle B=55°

Angle C=75°

Area=43,560 square feet

With that information, I need to find the length between A and B.

Answer 1

Let side AB lie horizontally with a 50 degree angle at vertex A and a 55 degree angle at vertex B. Let b represent the distance from A to B.

Picture extending the rays of those angles so that side AC intersects side BC(of course at a 75 degree angle) then drop an altitude from vertex C to side AB. Call the altitude h.

Let the point of intersection of the altitude with side AC be Q. Then AB = AQ + QB.

Now tan(50) = h/AQ and tan(55) = h/QB.

So, AQ=h/tan(50) and QB=h/tan(55).

The area of the triangle is 1/2 base times height.

43560 = 1/2 (AQ + QB)CQ = 1/2 (h/tan(50) + h/tan(55)) h

To solve for h (in steps):

h²/tan(50) + h²/tan(55) = 2(43560) : distribute the h and multiply both sides by 2.

Now multiply every term by tan(55)tan(50):

h²tan(55) + h²tan(50) = 2(43560)tan(55)tan(50)

Factor out h² on left side:

h²(tan(55) + tan(50)) = 2(43560)tan(55)tan(50)

h²=2(43560)tan(55)tan(50) / [tan(55) + tan(50)]

Take the square root of that mess to find h.

Then AQ = h / tan(50) and QB = h / tan(55)

Finally AB = h/tan(50) + h/tan(55)

Check it out here:

desmos.com/calculator/2h5hnuwjiw