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3 years ago

A 63 foot long cable is anchored in the ground and

Mathematics

1 answer:

myrzilka [38]3 years ago

8 0

Answer:

The height of the light pole to the nearest foot is approximately 50 feet

Step-by-step explanation:

The parameters given in the question are;

The length of the cable = 63 feet

The angle of elevation of the cable to the top of the light pole = 52°

With the assumption that the light pole is perpendicular to the ground, the figure formed by the light pole, the distance of the of the cable from the base of the light pole and the length of the cable form a right triangle

The question can be answered by using trigonometric relations for the sine of the given angle as follows;

$sin ( \theta) = \dfrac{Opposite \ leg \ length}{Hypotenuse \ length}$

The opposite leg length of the formed right triangle = The height of the light pole

The hypotenuse length = The length of the cable = 63 feet

The angle, θ = The angle of elevation = 52°

Plugging in the values, gives;

$sin (52 ^ {\circ}) = \dfrac{The \ height \ of \ the \ light \ pole }{63 \ feet}$

∴ The height of the light pole = 63 feet × sin(52°) ≈ 49.645 feet

The height of the light pole to the nearest foot ≈ 50 feet.

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Answer:

See below

Step-by-step explanation:

3. What are two ways that a vector can be represented?

Considering a vector $\vec{v}$ in some vector space $\mathbb R^n$ we have

$\vec{v} = \langle a,b\rangle$

This is the component form. I don't like that way. It is probably used in high school, but

$\vec{v} = \begin{pmatrix} a\\ b\\ \end{pmatrix}$

is preferable because the inner product on $\mathbb R^n$ is defined to be

$\langle a,b\rangle := \sum_{i = 1}^n a_i b_i$

You can also write it using linear form such as $\vec{v} = 2i+2j$

For this question, I think you meant

vectors

$\vec{u_1} = (-8, 12)$

$\vec{u_2} = (13, 15)$

Once

$\cos(\theta)=\dfrac{\vec{u_1} \cdot\vec{u_2}}{||\vec{u_1}||||\vec{u_2}||}$

Considering that the dot product is

$\vec{u_1}\cdot \vec{u_2} = (-8)\cdot 13 + 12\cdot 15 = -104+180= 76$

and the norm of $\vec{u_1}$ is $||\vec{u_1}|| = \sqrt{(-8)^2 + 12^2} = \sqrt{64 + 144}= \sqrt{208}$

and the norm of $\vec{u_2}$ is $||\vec{u_2}|| = \sqrt{13^2 + 15^2} = \sqrt{169 + 225}= \sqrt{394}$

Thus,

$\cos(\theta)=\dfrac{76}{\sqrt{208} \sqrt{394}} = \dfrac{19}{\sqrt{13}\sqrt{394}}=\dfrac{19}{\sqrt{5122}}$

$\therefore \theta = \arccos \left(\dfrac{19}{\sqrt{5122}} \right)$

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