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3 years ago

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The sum of the measures of angles for a triangle is 180°. Find the value of x and then the measures of the 3 angles

1 answer:

dmitriy555 [2]3 years ago

6 0

Answer:

The value of x is 50

The measures of the 3 angles are 50°, 100°, 30°

Step-by-step explanation:

Let us solve the question

∵ The sum of the measures of the angles of a triangle is 180°

∵ The measures of the 3 angles of the given Δ are x°, 2x°, (x - 20)°

→ Add them and equate the sum by 180

∴ x° + 2x° + (x - 20)° = 180°

→ Add the like terms

∵ (x + 2x + x) + (-20) = 180

∴ 4x - 20 = 180

→ Add 20 to both sides

∵ 4x - 20 + 20 = 180 + 20

∴ 4x = 200

→ Divide both sides by 4 to find x

∴ x = 50

∴ The value of x is 50

To find the measures of the 3 angles substitute x by 50 in their expressions

∵ The measure of one angle is x

∴ The measure of the 1st angle is 50°

∵ The measure of the 2nd angle is 2x

∴ The measure of the 2nd angle = 2(50)

∴ The measure of the 2nd angle is 100°

∵ The measure of the 3rd angle is (x - 20)

∴ The measure of the 3rd angle = (50 - 20)

∴ The measure of the 3rd angle is 30°

∴ The measures of the 3 angles are 50°, 100°, 30°

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Answer:

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Step-by-step explanation:

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Divide both sides by $\sqrt{3}$ :

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The first ratio I have can be found using $\frac{\pi}{6}$ in the first rotation of the unit circle.

The second ratio I have can be found using $\frac{7\pi}{6}$ you can see this is on the same line as the $\frac{\pi}{6}$ so you could write $\frac{7\pi}{6}$ as $\frac{\pi}{6}+\pi$ .

So this means the following:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

is true when $x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

where $n$ is integer.

Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.

So now we have a linear equation to solve:

$x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

Add $\frac{\pi}{8}$ on both sides:

$x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi$

Find common denominator between the first two terms on the right.

That is 24.

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$x=\frac{7\pi}{24}+n \pi$ (So this is for all the solutions.)

Now I just notice that it said find all the solutions in the interval $[0,2\pi)$ .

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So recall $0\le x$ .

Then $0 \le u+\frac{\pi}{8}$ .

Subtract $\frac{\pi}{8}$ on both sides:

$-\frac{\pi}{8}\le u$

Simplify:

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$-\frac{\pi}{8}\le u$

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$\tan(u)=\frac{1}{\sqrt{3}}$ with the condition:

$-\frac{\pi}{8}\le u$

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So now adding $\frac{\pi}{8}$ to both gives us the solutions to:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$ in the interval:

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The solutions we are looking for are:

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Let's simplifying:

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