Question

In any triangle ABC,prove that acos B cosC +b cos C cos A + c cos A cos B

Answer 1

Projection formulae is the length of any side of a triangle is equal to the sum of the projections of other two sides on it.

In Any Triangle ABC,
(i) a = b cos C + c cos B
(ii)  b = c cos A + a cos C
(iii) c = a cos B +  b cos A

Proof:   
Let ABC be a triangle. Then the following three cases arises:
Case I: If ABC is an acute-angled triangle then we get,
           a = BC = BD + CD ………………………… (i)Now from the triangle ABD we have,
cos B = BD/AB   
⇒ BD = AB cos B
⇒ BD = c cos B, [since, AB = c]
Again, cos C = CD/AC  
⇒ CD = AC cos C 
⇒ CD = b cos C, [since, AC = b]



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Now, substitute the value of BD and CD in equation (i) we get,
        a = c cos B + b cos C         
Note: We observe in the above diagram BD and CD are projections of AB and AC respectively on BC.

Case II: If ABC is an acute-angled triangle then we get,
            a = BC = CD - BD                                           ………………………… (ii)

Now from the triangle ADC we have,
cos C =  CD/AC   
⇒ CD = AC cos C
⇒ CD = b cos C, [since, AC = b]
Again, cos (π - B) = BD/AB  
⇒ BD = AB cos (π - B)



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⇒ BD = -c cos B, [since, AB = c and cos (π - θ) = -cos θ]
Now, substitute the value of BD and CD in equation (ii) we get,
       a = b cos C - (-c cos B)
⇒ a = b cos C + c cos B