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Mamont248 [21]
2 years ago
11

(a) If θ is in standard position, then the reference angle θ is the acute angle formed by the terminal side of θ and the . So th

e reference angle for θ = 110° is θ = °, and that for θ = 210° is θ =

Physics
1 answer:
noname [10]2 years ago
7 0

Answer:

70 degree, 30 degree

Explanation:

We are given that \theta is in standard position.

The reference angle \theta is the acute angle.

Acute angle :The angle is always less than 90 degree.

The reference angle formed by terminal side of angle \theta with x-axis.

When \theta=110^{\circ}

Then, the reference angle =\bar{\theta}=180-110

\bar{\theta}=70^{\circ}

When \theta=210^{\circ}

Therefore, the reference angle

\bar{\theta}=210-180

\bar{\theta}=30^{\circ}

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3 years ago
1. An airplane flies 800 kilometers east from Denver to Wichita in one hour. Then the plane flies west back to Denver in the sam
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B. The velocity of the second flight is negative compared to the speed.

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3 years ago
Two automobiles are 150 kilometers apart and traveling toward each other. One automobile
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8 0
2 years ago
a missile is moving 1810 m/s at a 20.0 degree angle. it needs to hit a target 19,500 m away in a 32.0 degree direction in 9.20 s
Ket [755]

Answer:

112 m/s², 79.1°

Explanation:

In the x direction, given:

x₀ = 0 m

x = 19,500 cos 32.0° m

v₀ = 1810 cos 20.0° m/s

t = 9.20 s

Find: a

x = x₀ + v₀ t + ½ at²

19,500 cos 32.0° = 0 + (1810 cos 20.0°) (9.20) + ½ a (9.20)²

a = 21.01 m/s²

In the y direction, given:

y₀ = 0 m

y = 19,500 sin 32.0° m

v₀ = 1810 sin 20.0° m/s

t = 9.20 s

Find: a

y = y₀ + v₀ t + ½ at²

19,500 sin 32.0° = 0 + (1810 sin 20.0°) (9.20) + ½ a (9.20)²

a = 109.6 m/s²

The magnitude of the acceleration is:

a² = ax² + ay²

a² = (21.01)² + (109.6)²

a = 112 m/s²

And the direction is:

θ = atan(ay / ax)

θ = atan(109.6 / 21.01)

θ = 79.1°

5 0
2 years ago
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