Question

Given: ABC is a right triangle with right angle C. AC=15 centimeters and m∠A=40∘ . What is BC ? Enter your answer, rounded to the nearest tenth, in the box

Answer 1

Answer:

12.6

Step-by-step explanation:

The other answer next to mine has a great explanation, except it says round to the nearest tenth, not the nearest whole number. The correct answer would be 12.6.

Answer 2

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 $90\°$.

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 $\alpha$:

• The side opposite to the right angle is called The Hypotenuse (h)

• The side opposite to the angle $\alpha$ is called the Opposite (O)

• The side next to the angle $\alpha$ is called the Adjacent (A)

So, going back to the triangle of our question (first figure):

• The Hypotenuse is AB

• The Opposite is BC

• The Adjacent is AC

Now, if we want to find the length of each side of a right triangle, we have to use the Pythagorean Theorem and Trigonometric Functions:

Pythagorean Theorem

$h^{2}=A^{2} +O^{2}$    (1)  

Trigonometric Functions (here are shown three of them):

Sine: $sin(\alpha)=\frac{O}{h}$    (2)

Cosine: $cos(\alpha)=\frac{A}{h}$    (3)

Tangent: $tan(\alpha)=\frac{O}{A}$   (4)

In this case the function that works for this problem is cosine (3), let’s apply it here:

$cos(40\°)=\frac{AC}{h}$    

$cos(40\°)=\frac{15}{h}$    (5)

And we will use the Pythagorean Theorem to find the hypotenuse, as well:

$h^{2}=AC^{2}+BC^{2}$    

$h^{2}=15^{2}+BC^{2}$    (6)

$h=\sqrt{225+BC^2}$   (7)

Substitute (7) in (5):

$cos(40\°)=\frac{15}{\sqrt{225+BC^2}}$    

Then clear BC, which is the side we want:

${\sqrt{225+BC^2}}=\frac{15}{cos(40\°)}$

${{\sqrt{225+BC^2}}^2={(\frac{15}{cos(40\°)})}^2$

$225+BC^{2}=\frac{225}{{(cos(40\°))}^2}$

$BC^2=\frac{225}{{(cos(40\°))}^2}-225$

$BC=\sqrt{158,41}$

$BC=12.58$

Finally $BC$ is approximately 13 cm