# A triangle has sides A, B, and C. The angle between sides A and B is pi/4 and the angle between sides B and C is pi/12. If side B has a length of 15, what is the area of the triangle?

Feb 15, 2016

The area of a triangle is given by $A r e a = \frac{1}{2} b h$ where $b$ is the length of the base and $h$ is the perpendicular height. Draw a diagram and calculate as shown below that the area is $88.35$ $c {m}^{2}$.

#### Explanation:

Drawing a clear diagram is essential for this kind of problem. Mine is not perfectly to scale, but it allows me to have clear in my own mind what information I have and what I'm trying to achieve.

(I've placed the image at the bottom of the page because the formatting works better that way.)

We know that the area of a triangle is given by:

$A r e a = \frac{1}{2} b h$

We already know that the base is $15$ $c m$, but we don't know the perpendicular height. I have constructed a right angled triangle, so we could use trig, but we don't know how far along the $15$ $c m$ side the added line hits.

We could use sine rule or cosine rule in the main triangle (they work in all triangles, not just right-angled ones), but we still need a little more information.

In radians, the angles in a triangle add up to $\pi$ (or ${180}^{o}$, if we worked in degrees). We already have $\angle A B$=$\frac{\pi}{4} = \frac{3 \pi}{12}$ and $\angle B C$=$\frac{\pi}{12}$ for a total of $\frac{4 \pi}{12}$, so $\angle A C$ (all of it) must be $\frac{8 \pi}{12}$.

Now we can use the sine rule to find the length of $A$ or $C$ or both:

$\frac{C}{\sin} c = \frac{B}{\sin} c \to C = B \sin \frac{c}{\sin} b = 15 \frac{\sin \left(\frac{\pi}{4}\right)}{\sin} \left(\frac{8 \pi}{12}\right) = 12.2$ $c m$

(for our purposes here, lowercase letters represent angles and uppercase represent sides, though the convention is usually the other way around)

Now that we know the length of $C$, we can use trig to find the height of the vertical line: the height of the triangle. Let's call that $' D '$:

$\cos \left(\frac{\pi}{12}\right) = \frac{D}{C} \to D = C \cos \left(\frac{\pi}{12}\right) = 12.2 \cos \left(\frac{\pi}{12}\right) = 11.78$ $c m$

Now we know the base and height of the triangle, we can calculate its area:

$A r e a = \frac{1}{2} b h = \frac{1}{2} \cdot 15 \cdot 11.78 = 88.35$ $c {m}^{2}$