Answer:
Explanation:
Right Triangles
The right triangles are those who have an internal angle of 90°. There are certain relations that only stand in right triangles, where the larger side is called the hypotenuse and the other two are the legs. The trigonometric relations in a right triangle are:
[tex]\displaystyle sin\alpha=\frac{y}{h}[/tex]
Where y is the opposite side to the angle [tex]\alpha[/tex] and h is the hypotenuse. We also have
[tex]\displaystyle cos\alpha=\frac{x}{h}[/tex]
In this formula, x is the adjacent side to [tex]\alpha[/tex]. Finally, we have
[tex]\displaystyle tan\alpha=\frac{y}{x}[/tex]
We cannot be sure what is the data and what are the results. We'll assume two given values: The base of the triangle x=6m and the angle adjacent to it [tex]\alpha=15^o[/tex]
The sine, cosine, and tangent of some angles called notable or special angles are widely known. Some of these angles are 0°, 30°, 45°, 60°, 90°.
The triangle of this problem has an angle of 15° which trigonometric functions are not so notable, but with some research we find
[tex]\displaystyle cos15^o=\frac{\sqrt{6}+\sqrt{2}}{4}[/tex]
[tex]\displaystyle tan15^o=2-\sqrt{3}[/tex]
With these values in mind, we can find the rest of the sides of the triangle. For example, to know the value of the hypotenuse, we set
[tex]\displaystyle tan15^o=\frac{x}{6m}[/tex]
Solving for x
[tex]x=6m\ tan15^o=6(2-\sqrt{3})=12-6\sqrt{3}[/tex]
Similarily
[tex]\displaystyle cos\alpha=\frac{6}{y}[/tex]
Solving for y
[tex]\displaystyle y=\frac{6}{cos\alpha}[/tex]
[tex]\displaystyle y=\frac{6}{\frac{\sqrt{6}+\sqrt{2}}{4}}[/tex]
[tex]\displaystyle y=\frac{24}{\sqrt{6}+\sqrt{2}}[/tex]
[tex]\displaystyle y=6(\sqrt{6}-\sqrt{2})[/tex]
