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Question
A man starts his morning walk at a point A reaches two points B and C and finally back to A such that ∠A = 60° and ∠B = 45°, AC = 4 km in the ∆ABC. Find the total distance he covered during his morning walk
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Solution
Given In ∆ABC
AC = 4 km
∠A = 60°,
∠B = 45°
∠C = 180° – (60° + 45°)
∴ ∠C = 180° – 105° = 75°
Using sine formula
`"BC"/sin"A" = "AC"/sin"B" = "AB"/sin "C"`
`"BC"/(sin 60^circ) = 4/(sin 45^circ) = "AB"/(sin 75^circ)`
`"BC"/(sin 60^circ) = 4/(sin 45^circ) = "AB"/(sin 75^circ)`
`"BC"/(sin 60^circ) = 4/(sin 45^circ)`
BC = `4/(sin 45^circ) xx sin 60^circ`
BC = `4/(1/sqrt(2)) xx sqrt(3)/2`
BC = `2sqrt(3) * sqrt(2)`
= `2sqrt(6)` km
`4/(sin 45^circ) = "AB"/(sin 75^circ)`
`4/(1/sqrt(2)) = "AB"/(sin(45^circ + 30^circ))`
`4sqrt(2) = "AB"/(sin 4^circ cos30^circ + cos 45^circ sin 30^circ)`
`4sqrt(2) = "AB"/(1/sqrt() * sqrt(3)/2 + 1/sqrt(2) * 1/2)`
`4sqrt(2) = "AB"/((sqrt(3) + 1)/(2sqrt(2))`
`4sqrt(2) = (2sqrt(2)"AB")/(sqrt(3) + 1)`
2 = `"AB"/(sqrt(3) + 1)`
AB = `2(sqrt(3) + 1)` km.
Total distance of morning walk
= AB + BC + AC
= `2(sqrt(3) + 1) + 2sqrt(6) + 4`
= `2sqrt(3) + 2 + 2sqrt(6) + 4`
= `2sqrt(3) + 6 + 2sqrt(6)`
= `6 + 2sqrt(6) + 2sqrt(3)` km.
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