Asked by mathstudent
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)
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)
Answers
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sweens
1-2sin^5B write in single trigonometric function
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