Posted by **mathstudent** on Friday, December 15, 2006 at 4:19pm.

There is an arbitrary triangle with angles A, B, and C and sides of lengths a, b, and c. Angle A is opposite side a.

How do I get the formulas:

b * cos C + c * cos B = a

c * cos A + a * cos C = b

a * cos B + b * cos A = c

Are these standard trig formulas? What are they called? Or where are they derived from?

thanks!

b * cos C + c * cos B = a

Draw the line at right angles to side a that splits the angle A (not necessarily in equal parts). Then, the triangle is slit in two right triangles, and you see that

a = a1 + a2

And:

a1 = b * cos C

a2 = c * cos B

Cyclically permuting a -- > b---> c ---> a (and the uppercase variables) give you the other two equations)

## Answer this Question

## Related Questions

- Geometry - x ________________<20* |<90* y | |<70* (hypotenuse length ...
- Pre-Calc - Consider a triangle ABC. C is a right angle, and side a is opposite ...
- geometry - If a triangle has sides of lengths a and b, which make a C-degree ...
- SAT math - Can someone please double check my true and false answers! 1. All ...
- Math - Yesterday I learned to solve a right triangle assuming I know the 3 ...
- Maths/Right Triangle - There must be a relationship (formula) between the ...
- math - A right triangle has a hypotenuse of length 3.00 m, and one of its angles...
- Math sin/cos - On a piece of paper draw and label a right triangle using the ...
- maths - shapes - Can you name these quadrilaterals? a) All sides are the same ...
- maths - Choose three options which are true: a) an angle of 150 degrees is ...

More Related Questions