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a_sh-v [17]
3 years ago
12

What angle relationships are created when parallel lines are intersected by a transversal?

Mathematics
1 answer:
natka813 [3]3 years ago
7 0

Answered by Mimiwhatsup: If two parallel lines are cut by a third line, the third line is called the transversal.

Relationships:

(1) Corresponding angles

(2) Vertically Opposite angles

(3) Alternate interior angles

(4) Alternate exterior angles

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Write each expression as an algebraic​ (nontrigonometric) expression in​ u, u > 0.
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Answer:

\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)=\frac{20\sqrt{u^2-100}}{u^2}\text{ where } u>0

Step-by-step explanation:

We want to write the trignometric expression:

\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)\text{ where } u>0

As an algebraic equation.

First, we can focus on the inner expression. Let θ equal the expression:

\displaystyle \theta=\sec^{-1}\left(\frac{u}{10}\right)

Take the secant of both sides:

\displaystyle \sec(\theta)=\frac{u}{10}

Since secant is the ratio of the hypotenuse side to the adjacent side, this means that the opposite side is:

\displaystyle o=\sqrt{u^2-10^2}=\sqrt{u^2-100}

By substitutition:

\displaystyle= \sin(2\theta)

Using an double-angle identity:

=2\sin(\theta)\cos(\theta)

We know that the opposite side is √(u² -100), the adjacent side is 10, and the hypotenuse is u. Therefore:

\displaystyle =2\left(\frac{\sqrt{u^2-100}}{u}\right)\left(\frac{10}{u}\right)

Simplify. Therefore:

\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)=\frac{20\sqrt{u^2-100}}{u^2}\text{ where } u>0

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Step-by-step explanation:

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