In triangle ABC, A= 50 degrees, B=10, and C=20 inches. Find a to nearest tenth

Question

I tried solving it but the calculator gave me a error and I would like help to see what I did wrong.

 

First i did 

the formula I used a^2=b^2+c^2-2BCCosA

50^2=10^2+20^2-(10)(20)CosA 

2500=100+400-400CosA

2500=500-400CosA

subtracted 500

2000=-400CosA

divided -400

2000/-400 = -400CosA/-400

then did

Cos^-1 (2000/-400=CosA)

Cos-1(2000/400)=A

Then I typed it on a calculator and got an error 

I'm not sure what I did wrong or I might have done the whole problem wrong. Any help would be appreciated 

Answer 1

In triangle ABC, A= 50 degrees, B=10, and C=20 inches. Find a to nearest tenth
I tried solving it but the calculator gave me a error and I would like help to see what I did wrong.

First i did
the formula I used a^2=b^2+c^2-2BCCosA
50^2=10^2+20^2-(10)(20)CosA ⇐ you erroneously put 50 for value a!  But a is what you are trying to find.The A that you were given is the measure of angle A, so the right side of this should say cos(50).

 

You have violated the rule, in math, of capitalizing B and C in your formula, when bc (coefficient of cosA, above) appear.In a triangle, A,B,C are the vertices, where angles live, and a, b, c are the lengths of the sides opposite those angles.

 

Answer 2

If A is 50 Degrees that would go in the "CosA" part of your equation. If b and c are 10in and 20 inches then you have put them in the correct part of the equation.

 

a^2 = 10^2+20^2 -2(10)(20)Cos(50)

 

That would be your equation. Try solving it from there!

Remember: a,b,c are the lengths of the sides and A,B,C are the angles across the sides.