With the usual notation in `DeltaABC` if
`angleA+angle B=120^(@),a=sqrt(3)+1 and b=sqrt(3)-1` then the ratio `angle A:angleB` is

Options

A. `7:1`
B. `3:1`
C. `9:7`
D. `5:3`

Answer

A

Explanation

For a `DeltaABC`, it is given that `a=sqrt(3)+1`
`b=sqrt(3)-1 and angle A+angle B=120^(@)`

Clearly, `angleC=60^(@) [ :. angleA+angleB+angleC=180^(@)]`
Now, by tangent law, we have
`"tan"(A-B)/(2)=(a-b)/(a+b)"cot"(C)/(2)`
`=((sqrt(3)+1)-(sqrt(3)-1))/((sqrt(3)+1)+(sqrt(3)-1))cot((60^(@))/2)`
`(2)/(2sqrt(3))"cot"(30^(@))`
`(1)/(sqrt(3))xxsqrt(3)=1`
`rArr tan ((A-B)/(2))=1=tan 45^(@)`
`rArr (A-B)/(2)=45^(@)`
`angle A-angleB=90^(@)`
On solving `angle-AangleB=90^(@) and angleA+angleB=120^(@)` we get,
`angleA=105^(@) and angleB=15^(@)`
`angleA:angleB =7:1`