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2 years ago

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To aid in sea navigation, Little Gull Island Lighthouse shines a light from a height of 91 feet above sea level with an unknown

angle of depression. If the beam of light shines on the sea surface at a point that is 865 feet away from the base of the lighthouse, what is the angle of depression?

1 answer:

algol132 years ago

7 0

Answer:

$\approx 6^\circ$

Step-by-step explanation:

Given that:

Little Gull Island Lighthouse shines a light from a height of 91 feet above the sea level.

The angle of depression is unknown.

Distance of the point at sea surface from the base of lighthouse is 865 ft.

This situation can be modeled or can be represented as the figure attached in the answer area.

The situation can be represented by a right angled $\triangle ABC$ in which we are given the base and the height of the triangle.

And we have to find the value of $\angle BAD \ or \ \angle C$ (Because they are the internal vertically opposite angles).

Using tangent ratio:

$tan\theta = \dfrac{Perpendicular}{Base}$

$tanC = \dfrac{AC}{BC}\\\Rightarrow tanC = \dfrac{91}{865}\\\Rightarrow tanC = 0.105\\\Rightarrow \angle C \approx 6^\circ$

Therefore, the angle of depression is: $\approx 6^\circ$

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$\frac{\sqrt[3]{16y^4}}{x^2}$

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The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

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From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

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