how in triangle ABC, b=acosC+ccosA
Answers
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: 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