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Eduardwww
1 month ago
4

What is the angle (in degrees) north of east from the post office to the grocery store in centralburg?

Mathematics
2 answers:
tester [12.3K]1 month ago
7 0

Applying SOH CAH TOA principles;

Definitions:

SOH stands for Sine = Opposite over Hypotenuse.

CAH indicates Cosine = Adjacent over Hypotenuse.

TOA denotes Tangent = Opposite over Adjacent.

Given Data:

Opposite side is 4 squares.

Adjacent side is 3 squares.

Angle ∅ is unknown.

Calculation:

Using the TOA rule based on the data:

Tan∅ = Opposite ÷ Adjacent

Tan∅ = 4 ÷ 3

Tan∅ = 1.33

∅ = Tan⁻¹ (1.33)

∅ = 53.13°

Final Answer:

∅ = 53.13°

Summary:

Using SOH CAH TOA rules to find the angle when opposite and adjacent sides are known resulted in ∅ = 53.13°. This method can be applied to similar problems. Thank you for reviewing, and now you can solve questions of this type.

zzz [12.3K]1 month ago
4 0
Using the SOH CAH TOA method:

Definitions:
SOH means Sine = Opposite ÷ Hypotenuse.
CAH means Cosine = Adjacent ÷ Hypotenuse.
TOA means Tangent = Opposite ÷ Adjacent.

Solution:

Given:
Opposite side = 4 blocks
Adjacent side = 3 blocks
Unknown angle ∅

Calculation:
We can find ∅ using TOA rule:

Tan∅ = Opposite ÷ Adjacent
Tan∅ = 4 ÷ 3
Tan∅ = 1.33
∅ = Tan⁻¹ (1.33)
∅ = 53.13°

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Answer:

0.40 feet

Step-by-step explanation:

For the first scenario, a 50-foot building casts a shadow of 1 foot. Let the angle of elevation of the sun from the shadow be denoted as θ.

Then:

Tan θ = \frac{opposite}{adjacent}

Tan θ = \frac{50}{1}

Tan θ = 50

⇒ θ = Tan^{-1} 50

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The elevation angle is roughly 88.85^{o}.

For a 20-foot pole,

Tan θ = \frac{opposite}{adjacent}

Tan 88.85^{o} = \frac{20}{x}

x = \frac{20}{Tan 88.85^{o} }

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The length of the pole's shadow is 0.40 feet.

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What is true about the solution of StartFraction x squared Over 2 x minus 6 EndFraction = StartFraction 9 Over 6 x minus 18 EndF
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Answer:

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Step-by-step explanation:

We have

\frac{x^2}{2x-6}=\frac{9}{6x-18}

Factor both sides' denominators.

\frac{x^2}{2(x-3)}=\frac{9}{6(x-3)}

Simplify.

\frac{x^2}{2}=\frac{9}{6}

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Verify

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\frac{\sqrt{3}^2}{2(\sqrt{3}-3)}=\frac{9}{6(\sqrt{3}-3)}

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18=18 ---> is valid.

Therefore,

x=-\sqrt{3} -----> is a legitimate solution.

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