Question

how in triangle ABC, b=acosC+ccosA​

Answer 1

Answer:

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

i hope this helps

Step-by-step explanation:

Answer 2

Answer: hence proved

LHS=RHS

b=b

Step-by-step explination:

In ΔABC b=acosC+ccosA

the cosine formula is cosA=      b²+c²-a²

                                                       2bc

b=acosC+ccosA.......(1)

we know the cosine formula

cosA=      b²+c²-a²

                    2bc

appling cosine formula in equation (1)

b=a(b²+a²-c²)  +c(c²+b²-a²)

          2ab            2bc

=b²+a²-c² + c²+b²-a²

    2b               2b

=2b²

 2b

=b

hence proved

LHS=RHS

b=b

solved