let the area of triangle `A B C` be `((sqrt(3)-1))/2,b=2` , and `c=(sqrt(3)-1)` , and `/_A` be acute. The measure of the angle `C` is `15^0` (b) `30^0` (c) `60^0` (d) `75^0`

Options

A. `15^(@)`
B. `30^(@)`
C. `60^(@)`
D. `75^(@)`

Answer

A

Explanation

Using `Delta = (1)/(2) bc sin A`, we get
`(1)/(2) xx 2 xx (sqrt3 -1) sin A = (sqrt3 -1)/(2)`
`:. sin A = (1)/(2) " or " A = 30^(@)`
`tan.(B -C)/(2) = (b-c)/(b+c) cot.(A)/(2)`
`= (3-sqrt3)/(sqrt3+1) xx (sqrt3+1)/(sqrt3-1) = sqrt3`
`rArr B -C = 120^(@)`
Also `B + C = 150^(@) rArr C = 15^(@)` `r = (Delta)/(s) = 1 rArr s = 6`
`:. (s-a) (s-b) (s-c) = s`
`rArr s^(3) -(a +b + c)s^(2) + (ab + bc + ca) s - abc = s`
`rArr 6^(3) -12 xx 36 + (ab + bc + ca) - 60 = 6`
`rArr ab + bc + ca = 47`
`:. (1)/(a) + (1)/(b) + (1)/(c) = (47)/(60) = 0.8` (nearly)