`triangleABC` के अज्ञात अंगो को निकालें, यदि `a = sqrt(3) +1, b = sqrt(3)-1, C = 60^(@)`

Explanation

हम जानते है कि त्रिभुज ABC में,
`cosC = (a^(2)+b^(2) - c^(2))/(2ab)`
या, `cos60^(@) = ((sqrt(3)+1)^(2) +(sqrt(3)-1)^(2)-c^(2))/(2(sqrt(3)+1)(sqrt(3)-1))`
या, `(1)/(2) = ((3+1+2sqrt(3)) + (3+1-2sqrt(3))-c^(2))/(2(3-1))`
या, `(1)/(2) = (8-c^(2))/(4), ` या, `2= 8 -c ^(2)` या, `c^(2) = 8 -2 = 6, therefore c = sqrt(6)`
फिर `(b)/(sinB) = (c)/(sinC) therefore (sqrt(3)-1)/(sinB) = (sqrt(6))/(sin60^(@)) = (sqrt(6))/((sqrt(3))/(2)) = 2sqrt(2)`.
`therefore sinB = (sqrt(3)-1)/(2sqrt(2)) = sin 15^(@)" " therefore B = 15^(@)`
`because A + B + C = 180^(@), " " therefore A + 15^(@) + 60^(@) = 180^(@)` अतः `A = 105^(@)`
दूसरी विधि : सूत्र `(A-B)/(2) = (a-b)/(a+b) cot""(C)/(2)`
`tan""(A-B)/(2) = ((sqrt(3)=1)-(sqrt(3)-1))/(sqrt(3)+1 +sqrt(3)-1) cot30^(@) = (1)/(sqrt(3)) =1 tan45^(@)`
`therefore (A -B)/(2) = 45^(@)` या, `A - B = 90^(@) " "...(1)`
साथ ही `A + B = 120^(@)`
(1) और (2) से,`A = 105^(@), B = 15^(@)`