In order to answer this question, the figure in the first picture will be helpful to understand what a right triangle is. Here, a right angle refers to
.
However, if we want to solve the problem we have to know certain things before:
In the second figure is shown a general right triangle with its three sides and another given angle, we will name it
:
- The side <u>opposite to the right angle</u> is called The Hypotenuse (h)
- The side <u>opposite to the angle
</u> is called the Opposite (O)
- The side <u>next to the angle
</u> is called the Adjacent (A)
So, going back to the triangle of our question (first figure):
Now, if we want to find the length of each side of a right triangle, we have to use the <u>Pythagorean Theorem</u> and T<u>rigonometric Functions:</u>
Pythagorean Theorem
(1)
Trigonometric Functions (here are shown three of them):
Sine:
(2)
Cosine:
(3)
Tangent:
(4)
In this case the function that works for this problem is cosine (3), let’s apply it here:
(5)
And we will use the Pythagorean Theorem to find the hypotenuse, as well:
(6)
(7)
Substitute (7) in (5):
Then clear BC, which is the side we want:
![{\sqrt{225+BC^2}}=\frac{15}{cos(40\°)}](https://tex.z-dn.net/?f=%7B%5Csqrt%7B225%2BBC%5E2%7D%7D%3D%5Cfrac%7B15%7D%7Bcos%2840%5C%C2%B0%29%7D)
![{{\sqrt{225+BC^2}}^2={(\frac{15}{cos(40\°)})}^2](https://tex.z-dn.net/?f=%7B%7B%5Csqrt%7B225%2BBC%5E2%7D%7D%5E2%3D%7B%28%5Cfrac%7B15%7D%7Bcos%2840%5C%C2%B0%29%7D%29%7D%5E2)
![225+BC^{2}=\frac{225}{{(cos(40\°))}^2}](https://tex.z-dn.net/?f=225%2BBC%5E%7B2%7D%3D%5Cfrac%7B225%7D%7B%7B%28cos%2840%5C%C2%B0%29%29%7D%5E2%7D)
![BC^2=\frac{225}{{(cos(40\°))}^2}-225](https://tex.z-dn.net/?f=BC%5E2%3D%5Cfrac%7B225%7D%7B%7B%28cos%2840%5C%C2%B0%29%29%7D%5E2%7D-225)
![BC=\sqrt{158,41}](https://tex.z-dn.net/?f=BC%3D%5Csqrt%7B158%2C41%7D)
![BC=12.58](https://tex.z-dn.net/?f=BC%3D12.58)
Finally
is approximately 13 cm