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Tju [1.3M]
3 years ago
14

How do you use special right triangles to solve for missing sides? (30-60-90 and 45-45-90 triangles)

Mathematics
1 answer:
Assoli18 [71]3 years ago
8 0

Answer: see below

Step-by-step explanation:

30 - 60 - 90 triangles have angles in the triangle measuring 30, 60, and 90 degrees. A 30 - 60 - 90 triangle also has special side ratios according to a side's location in the triangle.

The side across from the 30 degree angle is represented by x.

The side across from the 60 degree angle is represented by x\sqrt{3}.

The side across from the 90 degree angle is represented by 2x.

45 - 45 - 90 triangles have angles in the triangle measuring 45, 45, and 90 degrees. A 45 - 45 - 90 triangle has special side ratios similar to the 30 - 60 - 90 triangle.

The side across from either of the 45 degree angles is represented by x.

The side across from the 90 degree angle is represented by x\sqrt{2}.

These ratios can be used to find missing sides. If you know that a triangle is one of these special triangles and you also know one of its side lengths, you can plug the known length in for x in the proper place.

EX: you have a 30 - 60 - 90 triangle with a side length of 2 across from the 30 degree angle. You then know that the side across from 60 is 2\sqrt{3} and the side across from 90 is 4.

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If there are 16 boys and 48 girls in a room, fill out all of the possible ratios of boys to girls that could be made.
kondor19780726 [428]
These are all of the possible ratios:

- 1:3
- 8:24
-4:12
-2:6

Please vote my answer brainliest. thanks!
5 0
3 years ago
Three forces of 300 N in the direction of N30E, 400N in the direction of N60E and
andreyandreev [35.5K]

Split up each force into horizontal and vertical components.

• 300 N at N30°E :

(300 N) (cos(30°) i + sin(30°) j)

• 400 N at N60°E :

(400 N) (cos(60°) i + sin(60°) j)

• 500 N at N80°E :

(500 N) (cos(80°) i + sin(80°) j)

The resultant force is the sum of these forces,

∑ F = (300 cos(30°) + 400 cos(60°) + 500 cos(80°)) i

… … …  + (300 sin(30°) + 400 sin(60°) + 500 sin(80°)) j N

∑ F ≈ (546.632 i + 988.814 j) N

so ∑ F has a magnitude of approximately 1129.85 N and points in the direction of approximately N61.0655°E.

4 0
2 years ago
The measures of two complementary angles are m angle 1 = (10x+7) and m angle 2=(9x-12) find the measures of both angles
USPshnik [31]

Answer:

m<1 = 57°

m<2 = 33°

Step-by-step explanation:

To find the numerical measure of both angles, let's come up with an equation to determine the value of x.

Given that m<1 = (10x +7)°, and m<2 = (9x - 12)°, where both are complementary angles, therefore, it means, both angles will add up to give us 90°.

Equation we can generate from this, is as follows:

(10x + 7)° + (9x - 12)° = 90°

Solve for x

10x + 7 + 9x - 12 = 90

Combine like terms

19x - 5 = 90

Add 5 to both sides

19x = 90 + 5 (addition property not equality)

19x = 95

Divide both sides by 19

x = 5

m<1 = (10x +7)°

Replace x with 5

m<1 = 10(5) + 7 = 50 + 7 = 57°

m<2 = (9x - 12)

Replace x with 5

m<2 = 9(5) - 12 = 45 - 12 = 33°

5 0
3 years ago
What’s the equation of (6,39) and (-5,-49)
blondinia [14]
Y= 8x-9, because I need more characters
5 0
3 years ago
Find the equation for the line that passes through the point (1,-3) and that is parallel to the line with the equation 3/2x-2y=-
kondaur [170]

The equation for the line that passes through the point (1,-3) and that is parallel to the line with the equation \frac{3}{2}x - 2y = \frac{-17}{2} is:

y = \frac{3}{4}x - \frac{15}{4}

<h3><u>Solution:</u></h3>

Given that line that passes through the point (1, -3) and that is parallel to the line with the equation \frac{3}{2}x - 2y = \frac{-17}{2}

We have to find equation of line

<em><u>The slope intercept form is given as:</u></em>

y = mx + c

Where "m" is the slope of line and "c" is the y-intercept

Let us first find slope of line containing equation \frac{3}{2}x - 2y = \frac{-17}{2}

\frac{3}{2}x - 2y = \frac{-17}{2}

Rearrange the above equation into slope intercept form

\frac{3x}{2} + \frac{17}{2} = 2y\\\\y = \frac{3x}{4} + \frac{17}{4}

On comparing the above equation with slope intercept form y = mx + c,

m = \frac{3}{4}

So the slope of line containing equation \frac{3}{2}x - 2y = \frac{-17}{2} is m = \frac{3}{4}

We know that slopes of parallel lines are equal

So the slope of line parallel to line having above equation is also m = \frac{3}{4}

<em><u>Now let us find the equation of line having slope m = 3/4 and passes through point (1 , -3)</u></em>

Substitute m = \frac{3}{4} and (x, y) = (1 , -3) in slope intercept form

y = mx + c

-3 = \frac{3}{4}(1) + c\\\\c = -3 - \frac{3}{4}\\\\c = \frac{-15}{4}

<em><u>Thus the required equation of line is:</u></em>

substitute m = \frac{3}{4} and c = \frac{-15}{4} in slope intercept form

y = \frac{3}{4}x + \frac{-15}{4}\\\\y =\frac{3}{4}x - \frac{15}{4}

Thus the equation of line is found out

3 0
3 years ago
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