Question

A triangle has sides A, B, and C. The angle between sides A and B is `(pi)/2` and the angle between sides B and C is `pi/12`. If side B has a length of 24, what is the area of the triangle?

Answer 1

`A=77.172` `units^2`

Explanation:

`Acolor(white)(00)color(black)(pi/12)color(white)(0000000)acolor(white)(00)color(white)(0)`

`Bcolor(white)(00)color(white)(pi/1 3)color(white)(0000000)bcolor(white)(00)color(black)(24)`

`Ccolor(white)(00)color(black)(pi/2)color(white)(0000000)c color(white)(00)color(white)(0)`

Let's find the remaining angle, `B`:

`pi-pi/12-pi/2` leaves us with `(5pi)/12`

`Acolor(white)(00)color(black)(pi/12)color(white)(00000000)acolor(white)(00)color(white)(0)`

`Bcolor(white)(00)color(black)((5pi)/12)color(white)(.)color(white)(0000000)bcolor(white)(00)color(black)(24)`

`Ccolor(white)(00)color(black)(pi/2)color(white)(000000000)c color(white)(00)color(white)(0)`

Now we could use law of sines, but I noticed something... `pi/2` is a right angle! Here's what our triangle looks like:

`color(white)(a)color(white)(- - - - - - - - - -)color(black)(/)color(black)(|)`
`color(white)(a)color(white)(- - - - - - - - -)color(black)(/)color(white)(-)color(black)(|)`
`color(white)(a)color(white)(- - - - - - - -)color(black)(/)color(white)(- - .)color(black)(|)`
`color(white)(a)color(white)(- - - - - - -)color(black)(/)color(white)(- - -0)color(black)(|)`
`color(white)(- - - -)color(black)(c)color(white)(00000)color(black)(/)color(white)(- - - -)color(white)(/)color(black)(|)color(black)(a)`
`color(white)(a)color(white)(- - - - -)color(black)(/)color(white)(- - - - -0)color(black)(|)`
`color(white)(a)color(white)(- - - -)color(black)(/)color(white)(- - - - - -0)color(black)(|)`
`color(white)(a)color(white)(- - -)color(black)(/)color(white)(- - - - - - -0)color(black)(|)`
`color(white)(a)color(white)(- -)color(black)(/)color(white)(- - - - - - - -0)color(black)(|)`
`color(white)(-)color(black)(/)color(black)()color(black)(A)color(black)(...........................................)`
`color(white)(0000000000000000) b`

What we need to find is the height, or `color(red)(a)` and the base, or `color(red)(b)`

`Acolor(white)(00)color(black)(pi/12)color(white)(00000000)acolor(white)(00)color(white)(0)`

`Bcolor(white)(00)color(black)((5pi)/12)color(white)(.)color(white)(0000000)bcolor(white)(00)color(red)(24)`

`Ccolor(white)(00)color(black)(pi/2)color(white)(000000000)c color(white)(00)color(white)(0)`

To find length `color(red)(a)`, we should use `tan`:

`tan(A)=a/b`

`tan(pi/12)=a/24`

`tan(pi/12) xx 24 = a`

`a ~~ 6.431`

Now we can find the area:

`A=(a xx b)/2`

`A=(6.431 xx 24)/2`

`A=77.172` `units^2`