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

Answer 1

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.