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3 years ago

5

Navigators locate an object by measuring its location clockwise direction from due north. The measure of the angle is called the

bearing. Suppose a lighthouse's bearing is 135° from an island. Be sure to show and explain all work for each part. Part 1: Find the angle represented by the lighthouse on the unit circle. Part 2: Determine the exact ordered pair associated with the lighthouse on the unit circle. Part 3: Find the exact sine, cosine, and tangent of the angle on the unit circle.

1 answer:

kotegsom [21]3 years ago

5 0

Answer:

<em>Refer to attached.</em>

Part 1

Since bearing is taken from north, i.e. 90°, this will be represented by -45° on the units circle.

Part 2

The ordered pair is: (0.707, -0.707)

Part 3

sin (-45°) = -√2/2
cos (-45°) = √2/2
tan (-45°) = -1

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None of the expression are equivalent to $49^{(2t - 0.5)}$

Step-by-step explanation:

Given

$49^{(2t - 0.5)}$

Required

Find its equivalents

We start by expanding the given expression

$49^{(2t - 0.5)}$

Expand 49

$(7^2)^{(2t - 0.5)}$

$7^2^{(2t - 0.5)}$

Using laws of indices: $(a^m)^n = a^{mn}$

$7^{(2*2t - 2*0.5)}$

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This implies that; each of the following options A,B and C must be equivalent to $49^{(2t - 0.5)}$ or alternatively, $7^{(4t - 1)}$

A. $\frac{7^{2t}}{49^{0.5}}$

Using law of indices which states;

$a^{mn} = (a^m)^n$

Applying this law to the numerator; we have

$\frac{(7^{2})^{t}}{49^{0.5}}$

Expand expression in bracket

$\frac{(7 * 7)^{t}}{49^{0.5}}$

$\frac{49^{t}}{49^{0.5}}$

Also; Using law of indices which states;

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

$\frac{49^{t}}{49^{0.5}}$ becomes

$49^{t-0.5}}$

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B. $\frac{49^{2t}}{7^{0.5}}$

Expand numerator

$\frac{(7*7)^{2t}}{7^{0.5}}$

$\frac{(7^2)^{2t}}{7^{0.5}}$

Using law of indices which states;

$(a^m)^n = a^{mn}$

Applying this law to the numerator; we have

$\frac{7^{2*2t}}{7^{0.5}}$

$\frac{7^{4t}}{7^{0.5}}$

Also; Using law of indices which states;

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

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$7^{2t}\ *\ 7^{2*0.5}$

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Using law of indices which states;

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