Question

Please Answer 31st with explanation

Answer 1

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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]


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)



⇒ 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

Case III: If ABC is a right-angled triangle then we get,

             a = BC                                              ………………………… (iii)

and cos B =  BC/AB   

⇒ BC = AB cos B

⇒ BC = c cos B, [since, AB = c]

Now, substitute the value of BC in equation (iii) we get,

       a = c cos B

⇒ a = c cos B +C

⇒ a = c cos B + b cos C, [since C = 90° ⇒ cos C = cos 90 = 0] 

Therefore, in any triangle ABC we get, a = b cos C + c cos B

Similarly, we can prove that the formulae b = c cos A + a cos C and c = a cos B + b cos A.