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yaroslaw [1]
1 year ago
15

Consider a triangle ABC like the one below. Suppose that a = 31, b = 23, and c = 20. (The figure is not drawn to scale.) Solve t

he triangle. Carry your intermediate computations to at least four decimal places, and round your answers to the nearest tenth. If there is more than one solution, use the button labeled "or".​
Mathematics
1 answer:
krok68 [10]1 year ago
6 0

The solution to the triangle is A = 92.0, B = 47.9 and C = 40.1

<h3>How to solve the triangle?</h3>

The figure is not given;

However, the question can still be solved without it

The given parameters are:

a = 31, b = 23, and c = 20

Calculate angle A using the following law of cosine

a² = b² + c² - 2bc * cos(A)

So, we have:

31² = 23² + 20² - 2 * 23 * 20 * cos(A)

Evaluate

961 = 929 - 920 * cos(A)

Subtract 929 from both sides

32 =- 920 * cos(A)

Divide both sides by -920

cos(A) = -0.0348

Take the arc cos of both sides

A = 92.0

Calculate angle B using the following law of sine

a/sin(A) = b/sin(B)

So, we have:

31/sin(92) = 23/sin(B)

This gives

31.0189 = 23/sin(B)

Rewrite as:

sin(B) =23/31.0189

Evaluate

sin(B) =0.7415

Take arc sin of both sides

B = 47.9

Calculate angle C using:

C = 180 - 92.0 - 47.9

Evaluate

C = 40.1

Hence, the solution to the triangle is A = 92.0, B = 47.9 and C = 40.1

Read more about triangles at:

brainly.com/question/2217700

#SPJ1

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What is the equation of the line that passes through the points (4/5,1/5) and (1/2,3/2)?​
Setler [38]

Answer:

y-\frac{3}{2}=\frac{-13}{3}(x-\frac{1}{2}) point-slope form

13x+3y=11 (standard form)

Let me know if you prefer another form.

Step-by-step explanation:

The slope of a line can be found using \frac{y_2-y_1}{x_2-x_1} provided you are given two points on the line.

We are.

Now you can use that formula.  But I really love to just line up the points vertically then subtract them vertically then put 2nd difference over 1st difference.

 (4/5  ,  1/5)

-( 1/2  ,  3/2)

-----------------

3/10          -13/10

2nd/1st = \frac{\frac{-13}{10}}{\frac{3}{10}}=\frac{-13}{3} is our slope.

So the following is point-slope form for a linear equaiton:

y-y_1=m(x-x_1) \text{ where } m \text{ is slope and } (x_1,y_1) \text{ is a point on the line }    

Plug in a point (x_1,y_1)=(\frac{1}{2},\frac{3}{2}) \text{ and } m=\frac{-13}{3}.

This gives:

y-\frac{3}{2}=\frac{-13}{3}(x-\frac{1}{2})

I'm going to distribute:

y-\frac{3}{2}=\frac{-13}{3}x-\frac{-13}{6}

Now I don't like these fractions so I'm going to multiply both sides by the least common multiply of 2,3, and 6 which is 6:

6y-9=-26x+13

Add 26x on both sides:

26x+6y-9=13

Add 9 on both sides:

26x+6y=22 This is actually standard form of a line.

It can be simplified though.

Divide both sides by 2:

13x+3y=11 (standard form)

7 0
3 years ago
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