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mcr 3u1 number 3 section total determine the values of the trigonometri…

Question

mcr 3u1
number
3
section
total
determine the values of the trigonometric ratios for angles less than 360°; prove
simple trigonometric identities; and solve problems using the primary
trigonometric ratios, the sine law, and the cosine law;
/6
swer the following questions in the appropriate blanks.
) list the first negative co - terminal angle to 70°.
b) what is the related acute angle of 225°?
c) determine the principal angle of - 45°.
d) in which quadrant does the angle 189° terminate if the
angle is drawn in standard position?
e) in which quadrant does the angle - 210° terminate if
the angle is drawn in standard position?
f) what is the related acute angle of - 76°?

Explanation:

Part a)

Step1: Recall co - terminal angle formula

Co - terminal angles are given by \(\theta + n\times360^{\circ}\), where \(n\in\mathbb{Z}\). We want the first negative co - terminal angle of \(70^{\circ}\). Let \(n=- 1\).

$$70^{\circ}+(-1)\times360^{\circ}=70^{\circ}-360^{\circ}=-290^{\circ}$$

Part b)

Step1: Determine the quadrant of \(225^{\circ}\)

Since \(180^{\circ}<225^{\circ}<270^{\circ}\), \(225^{\circ}\) is in the third quadrant.

Step2: Use the formula for the related acute angle in the third quadrant

The formula for the related acute angle \(\alpha\) of an angle \(\theta\) in the third quadrant is \(\alpha=\theta - 180^{\circ}\).
\(\alpha = 225^{\circ}-180^{\circ}=45^{\circ}\)

Part c)

Step1: Recall the formula for the principal angle

The principal angle \(\theta_p\) of an angle \(\theta\) is given by \(\theta_p=\theta + 360^{\circ}\) if \(\theta<0\). For \(\theta=-45^{\circ}\), \(\theta_p=-45^{\circ}+360^{\circ}=315^{\circ}\)

Part d)

Step1: Use the quadrant range

First quadrant: \(0^{\circ}<\theta<90^{\circ}\), second quadrant: \(90^{\circ}<\theta<180^{\circ}\), third quadrant: \(180^{\circ}<\theta<270^{\circ}\), fourth quadrant: \(270^{\circ}<\theta<360^{\circ}\). Since \(180^{\circ}<189^{\circ}<270^{\circ}\), the angle \(189^{\circ}\) terminates in the third quadrant.

Part e)

Step1: Find the positive co - terminal angle

Let's find a positive co - terminal angle of \(-210^{\circ}\). Using \(\theta=-210^{\circ}+360^{\circ}=150^{\circ}\). Since \(90^{\circ}<150^{\circ}<180^{\circ}\), the angle \(-210^{\circ}\) (which is co - terminal with \(150^{\circ}\)) terminates in the second quadrant.

Part f)

Step1: Find the positive co - terminal angle

Find a positive co - terminal angle of \(-76^{\circ}\). Using \(\theta=-76^{\circ}+360^{\circ}=284^{\circ}\). Since \(270^{\circ}<284^{\circ}<360^{\circ}\), the related acute angle \(\alpha = 360^{\circ}-284^{\circ}=76^{\circ}\)

Answer:

a) \(-290^{\circ}\)
b) \(45^{\circ}\)
c) \(315^{\circ}\)
d) Third quadrant
e) Second quadrant
f) \(76^{\circ}\)