# 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?

Mar 2, 2018

$34.297 {\text{units}}^{2}$

#### Explanation:

We have a triangle that looks like:

Here, we have that the angle $C = \frac{\pi}{2}$, $b = 16 \text{units}$ and angle $A = \frac{\pi}{12}$.

Since all the angles of a triangle add up to $\pi$, angle $B = \frac{5 \pi}{12}$.

According to the sine rule:

$\sin \frac{A}{a} = \sin \frac{B}{b}$

We have to solve for $a$. Inputting:

$\frac{\sin \left(\frac{\pi}{12}\right)}{a} = \frac{\sin \left(\frac{5 \pi}{12}\right)}{16}$

$a = \frac{16 \cdot \sin \left(\frac{\pi}{12}\right)}{\sin \left(\frac{5 \pi}{12}\right)}$

$a = 4.287 \text{units}$

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 $\frac{1}{2} b h$ 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:

$A = \frac{1}{2} b h$

$A = \frac{1}{2} \cdot 16 \cdot 4.29$

$A = 34.297 {\text{units}}^{2}$

Use Method Number Two:

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

Again, for kicks, I'm doing both.

$\textcolor{w h i t e}{o \le o \le o \le}$Sub-method Number 1:

$\textcolor{w h i t e}{o \le o \le o \le}$We have the Pythagoras' Theorem ${a}^{2} + {b}^{2} = {c}^{2}$.

$\textcolor{w h i t e}{o \le o \le o \le}$We also know that $a = 4.29 , b = 16$. Inputting:

$\textcolor{w h i t e}{o \le o \le o \le}$${16}^{2} + {4.29}^{2} = {c}^{2}$

$\textcolor{w h i t e}{o \le o \le o \le}$${c}^{2} = 274.404$

$\textcolor{w h i t e}{o \le o \le o \le}$$c = 16.565 \text{units}$

Now for:

$\textcolor{w h i t e}{o \le o \le o \le}$Sub-Method Number 2

$\textcolor{w h i t e}{o \le o \le o \le}$We have the cosine rule: ${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cos C$

$\textcolor{w h i t e}{o \le o \le o \le}$We know the stuff we stated above, and $\textcolor{w h i t e}{o \le o \le o \le}$that $C = \frac{\pi}{2}$

$\textcolor{w h i t e}{o \le o \le o \le}$Since $\cos \left(\frac{\pi}{2}\right) = 0$, the cosine rule $\textcolor{w h i t e}{o \le o \le o \le}$$\textcolor{w h i t e}{o \le o \le o \le}$reduces to the Pythagoras' Theorem, which we $\textcolor{w h i t e}{o \le o \le o \le}$solved above, so skip it.

Now we go back to using Heron's Formula:

$A = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$, where $s = \frac{a + b + c}{2}$

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

$s = \frac{4.287 + 16 + 16.565}{2}$

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

$A = \sqrt{18.426 \left(18.426 - 4.287\right) \left(18.426 - 16\right) \left(18.426 - 16.565\right)}$

$A = \sqrt{1176.216}$

$A = 34.297 {\text{units}}^{2}$