Question
In any ΔABC, with usual notations, prove that `b^2=c^2+a^2-2ca cosB`.
Solution
Consider that for ΔABC,∠B is in a standard position i.e. vertex B is at the origin and the side BC is along positive x-axis. As ∠B is an angle of a triangle ∠B can be acute or B ∠B can be obtuse.
Using the Cartesian co-ordinate system in both figure (1) and figure (2)
we get B ≡ (0,0)A ≡ (c cos B, c sin B) and C ≡ (a,0)
Now consider l(CA) =b
`therefore b^2=(a-ccosB)^2+(0-csinB)^2` , by distance formula
`b^2=a^2-2ac cosB+c^2cos^2B+c^2sin^2B`
`b^2=a^2-2ac cosB+c^2(sin^2B+cos^2B)`
`b^2=a^2+c^2-2ac cosB`
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Solution In any ΔABC, with usual notations, prove that b2=c2+a2−2cacosB Concept: Trigonometric Functions - Solution of a Triangle.
