পুনরালোচনা: ক্ষেত্রফলের সাইন সূত্র

You have probably done problems concerning triangle areas that look like this: "Find the area of triangle ABC with base $a$ and height $h$." That's easy - you just use the formula \begin{equation} A = \frac{1}{2}ah \end{equation} to find the area.

But what if you weren't given the lengths of the triangle's base and height?

Furthermore, what if you were only given the lengths of two sides and the angle between the two sides?

No, don't start panicking. Fortunately, you can still use the normal area formula for finding the area with only those 3 pieces of information, but with an addition of basic trigonometry.

Using the image above, we can see that $$\begin{equation} \sin(C) = \frac{h}{b}, \end{equation}$$ which means that $$\begin{equation} h = b\sin(C). \end{equation}$$

The height can be expressed by a side and an angle! We can plug this back in our beloved triangle area formula: \begin{align} A &= \frac{1}{2}ah \
&= \frac{1}{2}a(b\sin(C))\
&= \frac{1}{2}ab\sin(C). \end{align}

Given two sides of a triangle $a$ and $b$ and the angle $C$ between sides $a$ and $b$ (also called the inclusive angle), the sine area formula gives that \begin{equation} A = \frac{1}{2}a(b\sin(C)). \end{equation}

We know that the area of a triangle $ABC$, here denoted as $[ABC]$, is equal to one half of the product of the base and the height.

A natural question to ask is, "What is the height?" Once we answer this question, we may find the area using only the sides and angles of the triangle.

Denote $AB=c$, $CA=b$, and $BC=a$. Drop a perpendicular from $C$ to $AB$; denote the foot of the altitude as $X$.

From the base-height formula, $$\left[\triangle ABC\right]=\frac{1}{2}c\cdot CX$$

Look at right triangle $CXA$. The hypotenuse is $CA=b$, and the angle opposite $CX$ is angle $CAX$. Thus, we see that $CX=b \sin A$. Even if angle $A$ is obtuse, $CX=b \sin(180^{\circ} -A)=b \sin A$, so we get the same result.

Plugging this result into our base-height formula, $$\left[\triangle ABC\right]=\frac{1}{2}bc\cdot \sin A$$

We can drop altitudes from $A$ and $B$ and obtain symmetric results. Thus, we have a formula for the area of a triangle: $$\left[\triangle ABC\right]=\frac{1}{2}bc\cdot \sin A=\frac{1}{2}ca\cdot \sin B=\frac{1}{2}ab\cdot \sin C$$

I'm pretty sure if you're here, you know one formula to calculate the area of a triangle. Let's just make sure you know it.

Alright, good. Now, let's use trig to find another cool formula for the area of a triangle. Let's draw a diagram:

In the diagram, $a = BC$, $b = AC$, and $c = AB$. We know that the area of the triangle is $$A = \frac{bh}{2} = \frac{BC \cdot AD}{2} = \frac{1}{2} \cdot BC \cdot AD$$

Now take a look at $\angle{C}$. Note that $\sin{C} = \frac{AD}{AC}$, which means $AC \cdot \sin{C} = AD$. We can substitute this into our above formula: $$A = \frac{1}{2} \cdot BC \cdot AD = \frac{1}{2} \cdot BC \cdot (AC \cdot \sin{C}) = \boxed{\frac{1}{2} \cdot ab\sin{C}}$$

We now have a new formula! The area of a triangle is given by half the product of the lengths of two sides of the triangle times the sine of the included angle.

Analogously, we can use the same method with the other two pairs of sides:

$$A = \frac{1}{2} \cdot ab\sin{C} = \frac{1}{2} \cdot bc\sin{A} = \frac{1}{2} \cdot ac\sin{B}$$

You will see eventually, that this can be used to derive the Law of Sines!