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 16, what is the area of the triangle?

Answer 1

`34.297"units"^2`

Explanation:

We have a triangle that looks like:

Here, we have that the angle `C=pi/2`, `b=16"units"` and angle `A=pi/12`.

Since all the angles of a triangle add up to `pi`, angle `B=(5pi)/12`.

According to the sine rule:

`sinA/a=sinB/b`

We have to solve for `a`. Inputting:

`(sin(pi/12))/a=(sin((5pi)/12))/16`

`a=(16*sin(pi/12))/(sin((5pi)/12))`

`a=4.287"units"`

Now there exist possibilities all around. There are a multitude of ways to solve for the area. Let's look at the two main ways of action.

• Take `a` as the height of the triangle and `b` the base, and use the formula `1/2bh` to solve for the area.
• Find `c` and use Heron's Formula to solve for the area.

Just for kicks, let's do both!

Use Method Number 1:

`A=1/2bh`

`A=1/2*16*4.29`

`A=34.297"units"^2`

Use Method Number Two:

We must find `c` first. To do this, too, we have two methods:

• Use the Pythagoras' Theorem
• Use the Cosine Rule

Again, for kicks, I'm doing both.

`color(white)(oleoleole)`Sub-method Number 1:

`color(white)(oleoleole)`We have the Pythagoras' Theorem `a^2+b^2=c^2`.

`color(white)(oleoleole)`We also know that `a=4.29, b=16`. Inputting:

`color(white)(oleoleole)` `16^2+4.29^2=c^2`

`color(white)(oleoleole)` `c^2=274.404`

`color(white)(oleoleole)` `c=16.565"units"`

Now for:

`color(white)(oleoleole)`Sub-Method Number 2

`color(white)(oleoleole)`We have the cosine rule: `c^2=a^2+b^2-2abcosC`

`color(white)(oleoleole)`We know the stuff we stated above, and `color(white)(oleoleole)`that `C=pi/2`

`color(white)(oleoleole)`Since `cos(pi/2)=0`, the cosine rule `color(white)(oleoleole)` `color(white)(oleoleole)`reduces to the Pythagoras' Theorem, which we `color(white)(oleoleole)`solved above, so skip it.

Now we go back to using Heron's Formula:

`A=sqrt(s(s-a)(s-b)(s-c))`, where `s=(a+b+c)/2`

Here, `a=4.287, b=16, c=16.565`. So:

`s=(4.287+16+16.565)/2`

`s=18.426`. Inputting all of that into Heron's Formula:

`A=sqrt(18.426(18.426-4.287)(18.426-16)(18.426-16.565))`

`A=sqrt(1176.216)`

`A=34.297"units"^2`

Two different methods, same answer!