Finding the Area of a Triangle Using Trigonometry

||A_\text{triangle} = \displaystyle \frac{\color{green}{a} \times \color{blue}{b} \times \sin \color{red}{C}}{2}||
Where

|\color{green}{a} \ \text{and} \ \color{blue}{b}| are measurements of the sides.

|\color{red}{C}| is the measurement of the angle formed by the sides, as follows:

To demonstrate the trigonometric formula, use the following triangle:

In this case, the following can be determined:
||A_\text{triangle} = \displaystyle \frac{\color{red}{a} \times h }{2}||
According to the medium right-angle triangle on the left, the following can be determined:
||\begin{align} \sin C &= \frac{h}{\color{green}{b}} \\
\Rightarrow \color{green}{b} \times \sin C &= h \end{align}||
Substituting the variable |h| gives:
||\begin{align} A_\text{triangle} &= \frac{\color{red}{a} \times h}{2}\\
&= \frac{ \color{red}{a} \times \color{green}{b} \times \sin C}{2}\end{align}||

A farmer has too much agricultural land and decides to sell off a portion of it.

What price should he charge for the portion of land if the going price is $15 / m2?

1. Identify the essential measurements.
   
   All the numbers essential for applying the trigonometric formula are given as follows:
   ||\begin{align} \color{green}{a} &= \color{green}{ 85 \ \text{m}}\\
   \color{blue}{b} &= \color{blue}{92 \ \text{m}}\\
   \text{m} \angle C &= 105^\circ \end{align}||

2. Apply the trigonometric formula.
   
   ||\begin{align}\color{red}{A_\text{triangle}} &= \frac{\color{green}{a} \times \color{blue}{b} \times \sin C}{2} \\
   &= \frac{\color{green}{85} \times \color{blue}{92} \times \sin 105^\circ}{2}\\
   &\approx 3 \ 776.77 \ \text{m}^2 \end{align}||

3. Interpret the answer.
   
   The selling price is | = 15 \times 3776.77 = \$56\ 651.55.|

Height inside the triangle
To calculate the area of ​​the following triangle, draw the height originating from vertex C:

Thus, the missing measurements can be found using the following trigonometric ratios:
||\begin{align} &1. &\sin 40^\circ &= \frac{\color{orange}{m \overline{CH}}}{\color{green}{m \overline{BC}}}\\
&&\Rightarrow \color{orange}{m \overline{CH}} &= \sin 40^\circ \times \color{green}{m \overline{BC}}\\\\&2.& \cos 40^\circ &= \frac{m \overline{BH}}{\color{green}{m \overline{BC}}} \\
&&\Rightarrow m \overline{BH} &= \cos 40^\circ \times \color{green}{m \overline{BC}}\\\\
&3.& \tan \color{red}{22^\circ} &= \frac{\color{orange}{m \overline {CH}}}{m \overline{AH}}\\
&&\Rightarrow m \overline{AH} &= \displaystyle \frac{\color{orange}{m \overline{CH}}}{\tan \color{red}{22^\circ}}\end{align}||
4. Considering |\overline{AB}| as the base and |\color{orange}{\overline{CH}}| as the height of the triangle, the area can be calculated as follows:
||\begin{align} \Delta ABC &= \frac{(m \overline{AH} + m \overline {BH}) \times \color{orange}{m \overline{CH}}}{2}\\
&= \frac{m\overline{AB} \times \color{orange}{m \overline{CH}}}{2}\end{align}||

The height outside the triangle
To calculate the area of ​​the following triangle, draw the height originating from vertex B:

Thus, the missing measurements can be found using the following trigonometric ratios:
||\begin{align} &1.&m \angle BCH &= 180^\circ - \color{blue}{m \angle BCA} \\
&&&= 180^\circ - \color{blue}{118^\circ}\\
&&&= 62^\circ\\\\
&2.& \sin \color{red}{22^\circ} &= \frac{\color{orange}{m \overline{BH}}}{\color{green}{m \overline{BA}}}\\
&& \Rightarrow \color{orange}{m \overline{BH}} &= \sin \color{red}{22^\circ} \times \color{green}{m \overline{BC}}\\\\
&3.& \color{red}{m ​\overline{AC}} &= m \overline {AH} - m \overline{CH}\end{align}||
According to |\Delta BCH|,
||\begin{align}\tan 62^\circ &= \frac{\color{orange}{m \overline {BH}}}{m \overline{CH}} \\ \Rightarrow m \overline {CH}
&= \frac{\color{orange}{m \overline{BH}}}{\tan 62^\circ}\end{align}||
According to |\Delta ABH|,
||\begin{align} \cos \color{red}{22^\circ} &= \frac{m \overline {AH}}{\color{green}{m \overline {AB}}} \\
m \overline {AH} &= \cos \color{red}{22^\circ} \times \color{green}{m \overline {AB}}\end{align}||
4. Considering |\color{red}{\overline {AC}}| as the base and |\color{orange}{\overline{BH}}| as the height, we can calculate the area of ​​triangle ABC,
||\begin{align} A_\Delta &= \frac{(m \overline {AH} - m \overline{CH}) \times \color{orange}{m \overline {BH}}}{2} \\
&= \frac{\color{red}{m \overline {AC}​} \times \color{orange}{m \overline{BH}}}{2}\end{align}||

Note: It is important that the right-angled triangle formed by drawing the height contains at least one angle measurement (other than the right angle) and at least one side measurement.

What is the area of the following triangle?

1. Draw the appropriate height

Thus,
||\begin{align}\text{m} \angle C &= 180^\circ - \color{blue}{70^\circ} - \color{green}{40^\circ} \\
&= 70^\circ\\
\Rightarrow \angle C &= \color{blue}{\text{m} \angle A} \\
&= 70^\circ\end{align}||

Therefore, |\Delta ABC | is an isosceles.

2. Use trigonometric ratios
   
   For the height, use the trigonometric ratio associated with the side |\overline {AH}| of |\Delta AHB| :
   ||\begin{align} \cos \color{blue}{70^\circ} &= \frac{\text{m} \overline{AH}}{15}\\
   \Rightarrow m \overline{AH} &= \cos \color{blue}{70^\circ} \times 15\\
   &\approx ​5.12 \ \text{cm}\\\\
   \sin \color{blue}{70^\circ} &= \frac{\color{orange}{m \overline{BH}}}{15} \\
   \Rightarrow \color{orange}{m \overline {BH}} &= \sin \color{blue}{70^\circ} \times 15\\
   &\approx \color{orange}{14.1 \ \text{cm}}\end{align}||
   
   In addition,
   ||\begin{align} m \angle BCA &= 180^\circ - \color{blue}{70^\circ} - \color{green}{40^\circ} \\
   &= 70^\circ\end{align}||
   
   Since we are dealing with an isosceles triangle,
   ||\begin{align} \color{red}{\text{m} \overline {AC}} &= 2 \times \text{m} \overline {AH} \\
   &= 2 \times 5.12 \\
   &= \color{red}{10.24\ \text{cm}}\end{align}||

3. Apply the formula
   
   ||\begin{align} A_\text{triangle}&= \frac{\color{red}{\text{base}} \times \color{orange}{\text{height}}}{2}\\
   ​​&= \frac{\color{red}{10.24} \times \color{orange}{14.1}}{2}\\
   &\approx 72.19 \ \text{cm}^2\end{align}||

4. Interpret the answer
   
   The area of the triangle is approximately 72.19 cm².

To establish a military base at a new location, the Generals from different squadrons sketch the following plan:

To fit all the soldiers and their equipment, the area of ​​the new base must be at least |25 \ \text{km}^2|. Does the plan above meet this requirement?

1. Draw the appropriate height
   
   To form two right-angled triangles, draw the following height:

2. Use trigonometric ratios
   
   To calculate the area of ​​the triangle, find |\color{blue}{\text{m} \overline {AB}}| .
   ||\begin{align}\sin 38^\circ &= \frac{\color{green}{\text{m} \overline {CH}}}{8 \ \text{km}} \\
   \color{green}{\text{m} \overline {CH}} &\approx 4.5 \ \text{km}\\\\
   \cos 38^\circ &= \frac{\text{m} \overline {AH}}{8 \ \text{km}} \\
   \text{m} \overline {AH} &\approx 6.62 \ \text{km}\\\\
   \tan 51^\circ &= \frac{\color{green}{\text{m} \overline {CH}}} {\text{m} \overline {BH}} \\
   \tan 51^\circ &= \frac{4.5 \ \text{km}}{\text{m} \overline {BH}} \\
   \text{m} \overline {BH} &\approx 4.36 \ \text{km}\\\\
   \color{blue}{\text{m} \overline{AB}} &= \text{m} \overline {AH} + \text{m} \overline {HB}
   \\
   &= 6.62 + 4.36 \\
   &= \color{blue}{10.98 \ \text{km}}\end{align}||

3. Calculate the area
   
   ||\begin{align} A_\text{triangle} &= \frac{\color{blue}{\text{base}} \times \color{green}{\text{height}}}{2}\\
   &= \frac { \color{blue}{m \overline{AB}}​ \times \color{green}{m \overline{CH}}}{2}\\
   &= \frac { \color{blue}{10.98}​ \times \color{green}{4.5}}{2}\\
   &\approx 24.71 \ \text{km}^2\end{align}||

4. Interpret the answer
   
   The Generals will have to go back to the drawing board, because the area of ​​the triangle they drew up is less than |25 \ \text{km}^2|.