পুনরালোচনা: 45-45-90 এবং 30-60-90 ত্রিভুজের ত্রিকোণমিতিক অনুপাত

Special Right Triangles

As you may know, there exist two particularly notable right triangles: the right triangle with acute angles measuring $45^{\circ}$ and $45^{\circ}$ (commonly known as a $45$-$45$-$90$ triangle for obvious reasons) and the right triangle with acute angles measuring $30^{\circ}$ and $60^{\circ}$ (commonly known as a $30$-$60$-$90$ triangle for similarly obvious reasons).

A $45$-$45$-$90$ triangle is always isosceles: two of its angles are congruent, so the sides opposite those angles are congruent as well.

Thus, because all such triangles are similar, they all have side lengths in the ratio $$1:1:\sqrt{2}.$$

A $30$-$60$-$90$ triangle is simply an equilateral triangle that has been cut in half down one of its altitudes. This means that the shortest side of the triangle is exactly half of the longest side.

Thus, because all such triangles are similar, they all have side lengths in the ratio $$1:\sqrt{3}:2.$$

Basics of Trigonometry

It is all very good to know the ratios of the side lengths of these two particular right triangles, but it would be even more helpful if we could extend this to all right triangles, regardless of angle measure.

We would also like a more concise way to express what we know about these right triangles, since it is rather tiresome to repeatedly say "a triangle with angles measuring $30^{\circ}, 60^{\circ},$ and $90^{\circ}$ has side lengths in the ratio of $1:\sqrt{3}:2$." How could we write that down mathematically?

Perhaps a function would help us.

Given an angle measure of one of the acute angles of a right triangle, one could construct a right triangle with that angle measure and find the ratios of sides in that triangle. Since there are three sides to choose from, one could pick a pair of sides in three different ways, so there are three basic functions of the angles that are ratios of sides.

We call these functions trigonometric functions (or, when speaking, "trig functions"), because they are the basis for the branch of mathematics known as trigonometry -- the study of the relationships between the sides and angles of triangles, which is exactly what we are doing now.

Trigonometric Functions

Before we define these three basic functions, as well as their reciprocals, we will put a name to each side.

For any acute angle on a right triangle, let us call the leg opposite the angle the opposite, and the leg adjacent to the angle the adjacent (Creative names, huh?). The third side is still the hypotenuse.

Now we can combine the sides in different ways to form ratios, and name each one (here, $\theta$ represents the acute angle we are using as our point of reference):

$$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}$$ $$\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}$$ $$\tan\theta = \frac{\text{opposite}}{\text{adjacent}}$$

Here, $\sin$ stands for the word sine, $\cos$ stands for the word cosine, and $\tan$ stands for the word tangent.

Taking their reciprocals, we find three more (but less commonly used, since they can easily be expressed as the reciprocals of the previous ones) trigonometric functions, completing the sextuplet:

$$\sec\theta = \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{1}{\cos\theta}$$ $$\csc\theta = \frac{\text{hypotenuse}}{\text{opposite}} = \frac{1}{\sin\theta}$$ $$\cot\theta = \frac{\text{adjacent}}{\text{opposite}} = \frac{1}{\tan\theta}$$

Here, $\sec$ stands for the word secant, $\csc$ stands for the word cosecant, and $\cot$ stands for the word cotangent.

(Creative names, huh? Satiate your curiosity for their etymologies here)

For example, we can find $\sin 30^{\circ}$ by finding the ratio of the leg opposite the $30^{\circ}$ angle to the hypotenuse in a $30$-$60$-$90$ triangle. By our work above, this is simply equal to $\frac{1}{2}$.

Let's try a few more examples (use your knowledge about the special triangles above; if your answer is not a whole number, express it as an exact decimal):

Congratulations! You now know the basic building blocks of trigonometry and can use them on $45$-$45$-$90$ and $30$-$60$-$90$ triangles.

There are two types of special right triangles: 45-45-90 and 30-60-90 right triangles.

45-45-90 Right Triangle

As you can tell from the name of the triangle, the $45$-$45$-$90$ right triangle has interior angles of $45^{\circ}$, $45^{\circ}$, and $90^{\circ}$.

Since two of the angles of this triangle are equal to $45^{\circ}$, the triangle is an isosceles triangle. Being a right triangle, this means that its two legs (the sides of the right angle) have equal length.

Suppose the length of each leg of the $45$-$45$-$90$ triangle is $a$ units.

By the Pythagorean Theorem, we can derive the length of the hypotenuse, denoted as $b$: $$\begin{align} b^2 & = a^2 + a^2 \qquad\\ & = 2a^2\\ \therefore b & = \sqrt{2}a\ \text{units} \end{align}$$

Therefore, the ratios of the sides of a $45$-$45$-$90$ right triangle can be expressed $a:a:b = 1:1:\sqrt{2}$.

30-60-90 Right Triangle

Similarly, the $30$-$60$-$90$ right triangle has interior angles of $30^{\circ}$, $60^{\circ}$, and $90^{\circ}$.

Unlike the 45-45-90 right triangle, the side ratios of the triangle are not as straightforward. We need to prove the ratio of the length of the shortest leg to the length of the hypotenuse before we are able to determine all the side ratios.

Let us create a diagram (below) in which we reflect the $30$-$60$-$90$ triangle along the longer leg ($AD$). Since the two $90^{\circ}$ angles fit together on the same line, we have created a larger triangle - that resembles an equilateral triangle!

We can easily prove that it is an equilateral triangle by noting our larger triangle's angles. One of the angles of our original $30$-$60$-$90$ right triangle is $60^{\circ}$, so there are two $60^{\circ}$ angles. The third angle is also $60^{\circ}$ because it is the sum of two $30^{\circ}$ angles.

Suppose the length of the shortest side ($CD$ or $BD$) of one $30$-$60$-$90$ triangle is $1$ unit. Therefore, the length of a side of the equilateral triangle would be $2$ units!