So I have another trigonometry question PLZ HELP ME I'm clueless at this rate I have like a little idea but I'm lost, SO PLZZZ!
I NEED IT BY JUNE 4
1. Lifeguard stands on a beach are approximately 300 ft apart. They are at 30-degree angles to a windsurfer on the water in between them. How far away is the windsurfer from each tower. Draw a diagram to help you picture the situation.
2. If the Earth is approximately 93 million miles from the sun, and Mercury is about 36 million miles from the sun, what are the possible distances between Earth and Mercury on a certain day if the angle between the sun and mercury is 22 degrees as measured from Earth?
1. Lifeguard stands on a beach are approximately 300 ft apart. They are at 30-degree angles to a windsurfer on the water in between them. How far away is the windsurfer from each tower. Draw a diagram to help you picture the situation.
2. If the Earth is approximately 93 million miles from the sun, and Mercury is about 36 million miles from the sun, what are the possible distances between Earth and Mercury on a certain day if the angle between the sun and mercury is 22 degrees as measured from Earth?
1 answer:
Answer:
1) The windsurfer is approximately 580 ft from each lifeguard stand.
2) The distance between the Earth and Mercury is approximately 61 million miles.
Step-by-step explanation:
The image other two situations described is presented in the attached image to this answer.
1) From the attached image, the windsurfer forms a right angled triangle with each of the lifeguard stand and the midpoint between the two lifeguard stands.
Hence, the angle at the top of the triangle is half of 30°, 15° and the distance from the midpoint of the lifeguard stands to the lifeguard stands is 300/2 = 150 ft.
Let the required distance of the windsurfer from each of the lifeguard stands be x.
Using trigonometric relations,
Sin 15° = (150/x)
x = 150 ÷ sin 15° = 150 ÷ 0.2588
x = 579.6 ft = 580 ft to the nearest whole number
2) From the image, the Sun, Earth and Mercury form a triangle.
Let the possible distance between the Earth and Mercury be y.
Using cosine rule,
y² = 36² + 93² - (2×36×93×cos 22°)
y² = 3,736.5769098208
y = √3,736.5769098208 = 61.13 million miles = 61 million miles to the nearest million miles.
Hope this Helps!!!
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To find the area of his exclusion zone you would need to understand that a triangle with dimensions of 3, 4, and 5 represent a right triangle.
This means the exclusion zone would be applied to the base and the height of the triangular space.
You would add 2 km to the 3 km, and 2 km to the 4 km to create a new height of 5 km and a new base of 6 km.
Please see the attached picture to understand this.
You will find the area of the total space created by the new triangle and subtract the space represented by the original triangle to find the area of the exclusion zone.
(1/2 x 6 x 5) - (1/2 x 4 x 3)
15 km² -6 km² equals 9 km².
The exclusion space is 9 km².
This means the exclusion zone would be applied to the base and the height of the triangular space.
You would add 2 km to the 3 km, and 2 km to the 4 km to create a new height of 5 km and a new base of 6 km.
Please see the attached picture to understand this.
You will find the area of the total space created by the new triangle and subtract the space represented by the original triangle to find the area of the exclusion zone.
(1/2 x 6 x 5) - (1/2 x 4 x 3)
15 km² -6 km² equals 9 km².
The exclusion space is 9 km².
Answer:
A) all real numbers greater than or equal to 2
Domain:
Step-by-step explanation:
The variable in the radical <em>x </em>cannot be less than 0 as that wouldn’t be defined. So, the maximum that <em>x </em>can be is 0, and . So, 0+2=2, and <em>y </em>is greater than or equal to 2.
Domain is x is greater than or equal to 0 because you cannot square a negative. So, it must not be less than 0. It could be 0 because square root of 0 is just 0.