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प्रश्न
Solve the triangle in which a = `(sqrt3 + 1)`, b = `(sqrt3 - 1)` and ∠C = 60°.
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उत्तर
Given: a = `(sqrt3 + 1)`, b = `(sqrt3 - 1)` and ∠C = 60°
By cosine rule,
c2 = a2 + b2 - 2ab cos C
`= (sqrt3 + 1)^2 + (sqrt3 - 1)^2 - 2(sqrt3 + 1)(sqrt3 - 1) cos 60^circ`
= 3 + 1 + `2sqrt3` + 3 + 1 - `2sqrt3` - 2(3 - 1)`(1/2)`
= 8 - 2 = 6
∴ c = `sqrt6` ....[∵ c > 0]
By sine rule,
`"a"/("sin A") = "b"/("sin B") = "c"/("sin C")`
∴ `(sqrt3 + 1)/("sin A") = (sqrt3 - 1)/("sin B") = sqrt6/("sin" 60^circ)`
∴ `(sqrt3 + 1)/("sin A") = (sqrt3 - 1)/("sin B") = sqrt6/(sqrt3//2) = 2sqrt2`
∴ sin A = `(sqrt3 + 1)/(2sqrt2) and sin "B" = (sqrt3 - 1)/(2sqrt2)`
∴ `sin "A" = sqrt3/(2sqrt2) + 1/(2sqrt2) and sin "B" = sqrt3/(2sqrt2) - 1/(2sqrt2)`
∴ sin A = `sqrt3/2 xx 1/sqrt2 + 1/2 xx 1/sqrt2`
∴ and sin B = `sqrt3/2 xx 1/sqrt2 - 1/2 xx 1/sqrt2`
∴ sin A = sin 60° cos 45° + cos 60° sin 45° and
sin B = sin 60° cos 45° - cos 60° sin 45°
∴ sin A = sin (60° + 45°) = sin 105°
and sin B = sin (60° - 45°) = sin 15°
∴ A = 105° and B = 15°
Hence, A = 105°, B = 15° and C = `sqrt6` units
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