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3 years ago

HELP GIVING BRAINIEST

Mathematics

1 answer:

iren [92.7K]3 years ago

3 0

Answer:

m<ACD = $50^{o}$

Step-by-step explanation:

From the question given, ΔACD is a right angled triangle. Then we can apply one of the properties of a triangle to it.

In the triangle ACD:

<ACD + <DAC + <ADC = 180 (sum of angles in a triangle)

<ACD + 40 + 90 = 180

<ACD + 130 = 180

<ACD = 180 - 130

<ACD = $50^{o}$

With the application of the property of the sum of interior angles of a triangle, the measure of <ACD is $50^{o}$ .

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wlad13 [49]

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saul85 [17]

Answer:

Step-by-step explanation:

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Equation of the line using point

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Hope this helps you.

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3 years ago

PLEASE HELP: <br><br><br> if √3 cos x - sin x = B cos(x+θ), show that b cosθ = √3 and b sin θ = 1

Lesechka [4]

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2 years ago

A post is supported by two wires (one on each side going in oppositedirections) creating an angle of 80° between the wires. The

Vladimir [108]

Using the Sine rule,

$\frac{\sin A}{A}=\frac{\sin B}{B}=\frac{\sin C}{C}$ $\begin{gathered} \text{Let A = 14m,} \\ Substituting the variables into the formula,Where the length of the wires are, AP = xm and BP = ym[tex]\begin{gathered} \frac{\sin80^0}{14}=\frac{\sin40^0}{x} \\ \text{Crossmultiply,} \\ x\times\sin 80^0=14\times\sin 40^0 \\ Divide\text{ both sides by }\sin 80^0 \\ x=\frac{14\sin40^0}{\sin80^0} \\ x=9.14m \end{gathered}$

Hence, the length of wire AP (x) is 9.14m.

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$A^0+P^0+B^0=180^0$ $\begin{gathered} \text{Where A}^0=\text{ unknown,} \\ P^0=80^0\text{ and,} \\ B^0=40^0 \\ A^0+80^0+40^0=180^0 \\ A^0+120^0=180^0 \\ A^0=180^0-120^0 \\ A^0=60^0 \end{gathered}$

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$\begin{gathered} \frac{\sin 60^0}{y}=\frac{\sin 80^0}{14} \\ \text{Crossmultiply,} \\ y\times\sin 80^0=14\times\sin 60^0 \\ \text{Didive both sides by }\sin 80^0 \\ y=\frac{14\times\sin 60^0}{\sin 80^0} \\ y=12.31m \end{gathered}$

Hence, the length of wire BP (y) is 12.31m

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1 year ago

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kati45 [8]

If you took a better picture I would be able to solve it. Sorry it is pretty blury

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3 years ago

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