Answer:
1. cos⁻¹(b/2a)-cos⁻¹(c/2a)
2.cos⁻¹(1/3)-cos⁻¹(2/3)
Step-by-step explanation:
Given:
r = a
AB = b
AC = c
To find : ∠BAC
Construction:
Join OA ,OB and OC.
Solution:
Now we have two triangles : ΔOAB and ΔOAC after construction.
We can find ∠OAB and ∠OAC by applying cosine formula for ΔOAB and ΔOAC.
∠BAC = ∠OAB - ∠OAC
Applying cosine formula for ΔOAB,
cos∠OAB = [tex]\frac{OA^{2}+AB^{2}-OB^{2}}{2.OA.AB}=\frac{a^{2}+b^{2}-a^2 }{2.a.b}=\frac{b}{2a}\\[/tex]
∠OAB = cos⁻¹(b/2a)
Applying cosine formula for ΔOAC,
cos∠OAC = [tex]\frac{OA^2+AC^2-OC^2}{2.OA.AC}=\frac{a^2+c^2-a^2}{2.a.c}=\frac{c}{2a}[/tex]
∠OAC = cos⁻¹(c/2a)
Now
∠BAC = ∠OAB - ∠OAC
= cos⁻¹(b/2a) - cos⁻¹(c/2a)
If a=3 b=2 c=4, we get
∠BAC = cos⁻¹(1/3) - cos⁻¹(2/3)
