prove that cotA+cotB+cotC=a²+b²+c²/4∆
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Answer:
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Answer:
Assume triangle ABC with a opposite A, b opposite B, and c opposite C.
Also assume you have already proved the formula for the area of a triangle given 2 sides and an included angle. Let K = area of triangle ABC. Then by this formula the following are true:
1/2 ab sinC = K
1/2 ac sinB = K
1/2 bc sinA = K
That is given two sides and an included angle the area is 1/2 the product of the two sides times the sine of the included angle.
Using the Law of Cosines we also have:
a2 = b2+c2-2bccosA
b2 = a2+c2-2accosB
c2 = a2+b2-2abcosC
The trick is to add these three equations together, simplify, divide by 4, and then replace 1/2 ac with K/sinB (for example).
a2+b2+c2=2a2+2b2+2c2-2bccosA-2accosB-2abcosC
This can be rewritten as:
a2+b2+c2=2bccosA + 2accosB + 2abcosC
Divide every term by 4:
I. (a2+b2+c2)/4 = 1/2 bc cosA + 1/2 ac cosB + 1/2 ab cosC
But 1/2 bc = K/sinA; 1/2 ac = K/sinB ; 1/2 ab = K/sinC (these from the earlier mentioned formulas for area)
Substitute into I giving on the right side: (K/sinA) cosA + (K/sinB) cosB + (K/sinC) cosC
Replace cos X / sin X with cot X and on right side we have:
= KcotA + KcotB + KcotC
Now divide by K (which is the area of the triangle)
II. (a2+b2+c2)/4K = cotA + cotB + cotC
Hence, Proved
Step-by-step explanation:
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