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icang [17]
3 years ago
5

Complete the assignment on a separate sheet of paper Please attach pictures of your work.

Mathematics
1 answer:
Irina18 [472]3 years ago
5 0

Answer:

<u>TO FIND :-</u>

  • Length of all missing sides.

<u>FORMULAES TO KNOW BEFORE SOLVING :-</u>

  • \sin \theta = \frac{Side \: opposite \: to \: \theta}{Hypotenuse}
  • \cos \theta = \frac{Side \: adjacent \: to \: \theta}{Hypotenuse}
  • \tan \theta = \frac{Side \: opposite \: to \: \theta}{Side \: adjacent \: to \: \theta}

<u>SOLUTION :-</u>

1) θ = 16°

Length of side opposite to θ = 7

Hypotenuse = x

=> \sin 16 = \frac{7}{x}

=> \frac{7}{x} = 0.27563......

=> x = \frac{7}{0.27563....} = 25.39568..... ≈ 25.3

2) θ = 29°

Length of side opposite to θ = 6

Hypotenuse = x

=> \sin 29 = \frac{6}{x}

=> \frac{6}{x} = 0.48480......

=> x = \frac{6}{0.48480....} = 12.37599..... ≈ 12.3

3) θ = 30°

Length of side opposite to θ = x

Hypotenuse = 11

=> \sin 30 = \frac{x}{11}

=> \frac{x}{11} = 0.5

=> x = 0.5 \times 11 = 5.5

4) θ = 43°

Length of side adjacent to θ = x

Hypotenuse = 12

=> \cos 43 = \frac{x}{12}

=> \frac{x}{12} = 0.73135......

=> x = 12 \times 0.73135.... = 8.77624.... ≈ 8.8

5) θ = 55°

Length of side adjacent to θ = x

Hypotenuse = 6

=> \cos 55 = \frac{x}{6}

=> \frac{x}{6} = 0.57357......

=> x = 6 \times 0.57357.... = 3.44145.... ≈ 3.4

6) θ = 73°

Length of side adjacent to θ = 8

Hypotenuse = x

=> \cos 73 = \frac{8}{x}

=> \frac{8}{x} = 0.29237......

=> x = \frac{8}{0.29237.....} = 27.36242..... ≈ 27.3

7) θ = 69°

Length of side opposite to θ = 12

Length of side adjacent to θ = x

=> \tan 69 = \frac{12}{x}

=> \frac{12}{x} = 2.60508......

=> x = \frac{12}{2.60508....}  = 4.60636.... ≈ 4.6

8) θ = 20°

Length of side opposite to θ = 11

Length of side adjacent to θ = x

=> \tan 20 = \frac{11}{x}

=> \frac{11}{x} = 0.36397......

=> x = \frac{11}{0.36397....}  =30.22225.... ≈ 30.2

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Hey there!

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  • Answer:

\green{\boxed{\red{\bold{\sf{y = \dfrac{7}{6}x - 3}}}}}

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To find the equation of a line, we first have to determine its slope knowing that parallel lines have the same slope.

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Assuming that we want to get the equation in Slope-Intercept Form, let's substitute m = 7/6:

Slope-Intercept Form:

\sf{y = mx + b} \\ \sf{Where \: m \: is \: the \: slope \: of \:  the \: line \: and \: b \: is \: the \: y-intercept.}

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We know that the coordinates of the point (0 , -3) verify the equation since it is on the line \: \sf{d_1} \:. Now, replace y with -3 and x with 0:

\implies \sf{\overbrace{-3}^{y} = \dfrac{7}{8} \times \overbrace{0}^{x} + b} \\ \\ \implies \sf{-3 = 0 + b} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \sf{\boxed{\bold{b = -3}} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:

Therefore, the equation of the line \: \bold{d_1} \: is \green{\boxed{\red{\bold{\sf{y = \dfrac{7}{6}x - 3}}}}}

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