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3 years ago

12

Help me please solves this my math teacher will yell at me if I don’t get this done!!!

1 answer:

VARVARA [1.3K]3 years ago

5 0

The answer is 75

First you subtract 180 and 165 to find that smallest angle inside the triangle(15). Then you add 15 and 90,the two angles you know, and do 180 - 105= 75.

The trick is that all angles inside a triangle equal 180 degrees. Hope this helps!

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Step-by-step explanation:

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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 th

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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 <u>opposite to the right angle</u> is called The Hypotenuse (h)

The side <u>opposite to the angle $\alpha$ </u> is called the Opposite (O)

The side <u>next to the angle $\alpha$ </u> 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 <u>Pythagorean Theorem</u> and T<u>rigonometric Functions:</u>

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

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