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exis [7]
4 years ago
12

Triangle ABC is a right triangle and sin(53o) = StartFraction 4 Over x EndFraction. Solve for x and round to the nearest whole n

umber.
Triangle A B C is shown. Angle A C B is a right angle and angle B A C is 53 degrees. The length of B C is 4 centimeters, the length of A C is y, and the length of hypotenuse A B is x.

Which equation correctly uses the value of x to represent the cosine of angle A?

cos(53o) = StartFraction 4 Over x EndFraction
cos(53o) = StartFraction y Over 5 EndFraction
cos(53o) = StartFraction x Over 4 EndFraction
cos(53o) = StartFraction 5 Over y EndFraction

Mathematics
1 answer:
zmey [24]4 years ago
4 0

Answer:

cos(53°)=StartFraction y Over 5 EndFraction

Step-by-step explanation:

see the attached figure to better understand the problem

step 1

Find the value of x

we know that

sin(53\°)=\frac{4}{x}

Solve for x

x=\frac{4}{sin(53\°)}

x=5

step 2

Find the cosine of angle A

The cosine of angle A is equal to divide the adjacent side to angle A (AC) by the hypotenuse (AB)

cos(53\°)=\frac{y}{x}

substitute the value of x

cos(53\°)=\frac{y}{5}

so

cos(53°)=StartFraction y Over 5 EndFraction

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Mandarinka [93]

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Step-by-step explanation:

Given

Line passes through (3,-1)\ \text{ and}\ (4,7)

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What’s the number between -6 and -7 on a number line
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Please help with 7, 9 and 11!! i am stuck on problems
Sholpan [36]

7) (x, TU, TB) = (4, 8, 13)

9) (x, RS, MN) = (8, 41, 41)

11) (AB, BC, AC) = (42, 42, 84)

<h3>What are the values of the variable x and the lengths of the line segments?</h3>

In this question we have three cases of colinear line segments, each of them can be solved by using definitions from Euclidean geometry and algebraic handling:

Case 7 - TU = 2 · x, UB = 3 · x + 1, TB = 21

TB = TU + UB

21 = 2 · x + 3 · x + 1

21 = 5 · x + 1

5 · x = 20

x = 4

Then, the lengths of each line segment are:

TB = 21

TU = 2 · 4

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The solutions are (x, TU, TB) = (4, 8, 13).

Case 9 - RS = 3 · x + 17, MN = 7 · x - 15

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3 · x + 17 = 7 · x  - 15

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And the lengths of each line segment are:

RS = MN = 41

The solutions are (x, RS, MN) = (8, 41, 41).

Case 11 - AB = 2 · x - 8, BC = x + 17

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And the lengths of each line segment are:

AB = 2 · 25 - 8

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BC = AB

BC = 42

AC = 42 + 42

AC = 84

The solutions are (AB, BC, AC) = (42, 42, 84).

To learn more on line segments: brainly.com/question/25727583

#SPJ1

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