Trigonometry Comparison of Right Triangles

Trigonometry is closely related to triangles. There are several types of Trigonometric Comparisons in Right Triangles, namely sin, cos, tan.

Hello Friends! This time we will learn about trigonometric ratios in right triangles. Let’s see more in here.

Trigonometry is closely related to triangles. There are several types of trigonometric comparisons, namely sin, cos, tan, secan (sec), cossec (csc), and cotangent (cot). Trigonometry (from Greek trigonon “triangle” + metron “measure”).

Trigonometry helps us find angles and distances, and is widely used in science, engineering, video games, and more. We encounter many things related to trigonometric comparisons. Trigonometric comparisons can be used to solve contextual problems related to angles and distances. By studying trigonometric ratios we can train reasoning in solving these problems.

Let’s start by remembering the right triangle first…

Right triangle

Look at the following right triangle image!

Right angles are indicated by small squares and suppose other angles are marked $\alpha$. So some of the side terms are:

• adjacent is the side adjacent (next to) angle $\alpha$,
• opposite which is the side opposite the corner $\alpha$,
• hypotenuse is the longest side of a right triangle.

Remember the Pythagorean theorem??? Yeahh… we will also use that in studying trigonometry…

In the right triangle image above, this applies

• $c^2=a^2+b^2$ or $c=\sqrt{a^2+b^2}$ to find the length of the hypotenuse or longest side or hypotenuse if the length of the right angle is known
• $a^2=c^2-b^2$ or $a=\sqrt{c^2-b^2}$ to find the length of the front side if the lengths of the hypotenuse and side sides are known
• $b^2=c^2-a^2$ or $b=\sqrt{c^2-a^2}$ to find the length of the side if the lengths of the hypotenuse and front side are known

Example of a Pythagorean question

Determine the value of x in the following triangle.

Solution

$x$ is not the longest side so you can use formula 2 or 3 $$x=\sqrt{7^2-2^2}\\x=\sqrt{49-4}\\x=\sqrt{45}$$ So, the value of x is $\sqrt{45}$.

Trigonometric Comparisons in Right Triangles

Note: Terms that often appear are sin, cos, tan… the easiest to memorize are SOH, CAH, TOA.. 

Example of Trigonometry Comparison Questions

1. Determine the trigonometric ratio values ​​$\sin\theta, \cos\theta, \tan\theta$ in the following figure

   Solution ✍️

   • find the length of side AC or side side with pythagoras $$\begin{align*} AC&=\sqrt{AB^2-BC^2}\\ &=\sqrt{10^2-8^2}\\ &=\sqrt{100-64}\\ &=\sqrt{36}\\ AC&=6 \end{align*}$$
   • mencari nilai $\sin\theta, \cos\theta, \tan\theta$ $$\sin\theta=\frac{de}{mi}=\frac{BC}{AB}=\frac{8}{10}=\frac{4}{5}$$ $$\cos\theta=\frac{sa}{mi}=\frac{AC}{AB}=\frac{6}{10}=\frac{3}{5}$$ $$\tan\theta=\frac{de}{sa}=\frac{BC}{AC}=\frac{8}{6}=\frac{4}{3}$$

2. It is known that the triangle PQR is right-angled at Q with side length $PR=\sqrt5$ units and side length $QR= 2$ units. If $\angle RPQ = \alpha$, determine the values ​​of $\sin\alpha, \cos\alpha, \tan\alpha$!

   Solution ✍️

   • draw a right triangle and Look at the following picture of a right triangle
   • find the length of side PQ or side side with Pythagoras. $$\begin{align*} PQ&=\sqrt{PR^2-RQ^2}\\ &=\sqrt{(\sqrt{5})^2-2^2}\\ &=\sqrt{5-4}\\ &=\sqrt{1}\\ PQ&=1 \end{align*}$$
   • looks for the values ​​$\sin\alpha, \cos\alpha, \tan\alpha$ $$\sin\alpha=\frac{de}{mi}=\frac{RQ}{PR}=\frac{2}{\sqrt{5}}=\frac{2}{5}\sqrt{5 }$$ $$\cos\alpha=\frac{sa}{mi}=\frac{PQ}{PR}=\frac{1}{\sqrt{5}}=\frac{1}{5}\sqrt{5}$$ $$\tan\alpha=\frac{de}{sa}=\frac{RQ}{PQ}=\frac{2}{1}=2$$

3. Given $\cos \beta^\circ = \frac{1}{2}$ and $\beta$ acute angle ($0^\circ < \beta < 90^\circ$). Find the value of the other trigonometric ratio of angle $\beta$.

   Solution ✍️

   • identify the sides of the triangle $\cos \beta^\circ = \frac{1}{2}=\frac{sa}{mi}$, we get the length of side 1 and the length of hypotenuse 2
   • draw a right triangle, for example $\triangle ABC$ is right angled at B and $\angle BAC = \beta$ then AB=1 and AC=2 Look at the following image
   • find the length of side AB or front side using Pythagoras. $$\begin{align*} BC&=\sqrt{AC^2-AB^2}\\ &=\sqrt{2^2-1^2}\\ &=\sqrt{4-1}\\ BC&=\sqrt{3}\end{align*}$$
   • looks for the values ​​$\sin\beta, \cos\beta, \tan\beta$ $$\sin\beta=\frac{de}{mi}=\frac{BC}{AC}=\frac{\sqrt{3}}{2}=\frac{1}{2}\sqrt{3 }$$ $$\cos\beta=\frac{sa}{mi}=\frac{AB}{AC}=\frac{1}{2}$$ $$\tan\beta=\frac{de}{sa}=\frac{BC}{AB}=\frac{\sqrt{3}}{1}=\sqrt{3}$$

4. Given $\tan \alpha^\circ = \dfrac{3}{4}$ and $\alpha$ acute angle ($0^\circ < \alpha < 90^\circ$). determine the value of $\dfrac{\sin\alpha-\tan\alpha}{\sin^2\alpha+\cos^2\alpha} $.

   Solution ✍️

   • identify the sides of the triangle $\tan \alpha^\circ = \dfrac{3}{4}= \dfrac{de}{sa}$, we get the length of the front side is 3 and the length of the side is 4
   • draw a right triangle, for example $\triangle ABC$ is right angled at B and $\angle BAC = \alpha$ then AB=4 and BC=3 Look at the following image
   • find the length of side AC or hypotenuse using Pythagoras. $$\begin{align*} AC&=\sqrt{AB^2+BC^2}\\ &=\sqrt{3^2+4^2}\\ &=\sqrt{9+16}\\ AC&=\sqrt{25}=5 \end{align*}$$
   • finding value $\dfrac{\sin\alpha-\tan\alpha}{\sin^2\alpha+\cos^2\alpha}$ $$\begin{align*} \frac{\sin\alpha-\tan\alpha}{\sin^2\alpha+\cos^2\alpha}&=\frac{\frac{3}{5}-\frac{3}{4}}{\left(\frac{3}{5}\right)^2+\left(\frac{4}{5}\right)^2}\\ &=\frac{\frac{12}{20}-\frac{15}{20}}{\frac{9}{25}+\frac{16}{25}}\\ &=\frac{-\frac{3}{20}}{\frac{25}{25}}\\ &=-\frac{3}{20} \end{align*}$$ So, value $\dfrac{\sin\alpha-\tan\alpha}{\sin^2\alpha+\cos^2\alpha}=-\dfrac{3}{20}$