In triangle ABC prove that a^2 *cot A + b^2 *cot B + c^2 *cot C =4*Delta. where Delta=area of the triangle.
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Expert's answer
2014-12-29T13:03:09-0500
Answer on Question #50171 - Math - Trigonometry
In triangle ABC prove that a2⋅cotA+b2⋅cotB+c2⋅cotC=4⋅Δ , where Δ - area of the triangle.
Solution:
O is the center of the circumcircle. The area of the triangle AOB is given by the following formula
SAOB=21AO⋅OB⋅sinAOB
Since O is the center of the circumcircle, the angle AOB two times larger than angle C of the triangle ABC . Using this and the facts that AO and OB are the radii of the circumcircle (we'll denote its value by R in the further) and that sin2α=2sinα⋅cosα , the previous formula can be rewritten in the following way
SAOB=21AO⋅OB⋅sinAOB=21R2⋅sin2C=R2⋅sinC⋅cosC
The area of the triangle ABC can be given as a sum of areas of triangles AOB , BOC and COA :
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