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If the base of a triangle and the ratio ...

If the base of a triangle and the ratio of tangent of half of base angles are given, then identify the locus of the opposite vertex.

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In triangle ABC, base BC is given.
i.e., BC = a (constant)
Also, given that
`(tan.(B)/(2))/(tan.(C)/(2))=k("constant")`

`therefore" "(sin.(B)/(2)cos.(C)/(2))/(sin.(C)/(2)cos.(B)/(2))=k`
`rArr" "(sin.(B)/(2)cos.(C)/(2)-sin.(C)/(2)cos.(B)/(2))/(sin.(B)/(2)cos.(C)/(2)+sin.(C)/(2)cos.(B)/(2))=(k-1)/(k+1)`
`rArr" "(sin.(B-C)/(2))/(sin.(B+C)/(2))=(k-1)/(k+1)`
`rArr" "(2sin.(B-C)/(2)cos.(B+C)/(2))/(2sin.(B+C)/(2)cos.(B+C)/(2))=(k-1)/(k+1)`
`rArr" "(sinB-sinC)/(sin(B+C))=(k-1)/(k+1)`
`rArr" "(sinB-sinC)/(sinA)=(k-1)/(k+1)"[As in triangle sin (B + C) sin A]"`
`rArr" "(b-c)/(a)=(k-1)/(k+1)" (using sine rule)"`
`rArr" "b-c=a((k-1)/(k+1))`
`rArr" "AC-AB=a((k-1)/(k+1))" (= constant)"`
Thus, difference of two variable sides AC and AB is constant. Hence, locus of vertex A is hyperbola with B and C as foci.
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