Solve each problem. See Examples 1–4. Diameter of the Sun To det...

Question

Solve each problem. See Examples 1–4.  Diameter of the Sun To determine the diameter of the sun, an astronomer might sight with a transit (a device used by surveyors for measuring angles) first to one edge of the sun and then to the other, estimating that the included angle equals 32'.  Assuming that the distance d from Earth to the sun is 92,919,800 mi, approximate the diameter of the sun.

Accepted Answer

Welcome back. Everyone in this problem. We're talking about astrophysicists trying to approximate the diameter of the sun. There's an equipment called a total station that is used to measure angles. And let's say, hypothetically an astrophysicist used it on Mars to set the edges of the sun just as shown in the figure. The included angle was found to be 21 minutes. And we want to estimate the diameter of the sun if the distance X from Mars to the sun is 141 million, 610,500 miles for our answer choices. A is 865,000 miles. B is 721,000 miles. C is 912,000 miles and B is 798,000 miles. Now, to make sense of our problem, let's try to draw a diagram that can help us to see some of the length and, and what's really going on here. So notice that our diagram forms an Isosceles triangle. OK. And or Isosceles triangle that's formed is made up of two adjacent right triangles back to back. So let's call the peak of our triangle M and we have our right angles at on the basis of the triangle or on the base of our triangle? Ok. Now, what do we know? Well, we already know that the distance from Mars to the sun is the length X. So let's label that on a diagram as 141 million, 610,500 miles. What else do we know? We also know that the included anger was found to be 21 minutes. So if that included anger, that is the anger from one edge of the sun to the other is 21 minutes. Ok. Then one of the angles in the right triangle would be a half of 21 minutes. Now, what else do we know? Well, we know that we're trying to estimate the diameter of the sun. So on our diagram, the diameter of the sun would be the base of the Isosceles triangle. Thus, that tells us that the base of one of the right triangles would be a half of the diameter or the radius of the sun. So if we can figure out the radius of the sun, then we should be able to figure out its diameter. And we can do that because notice we have an angle in one of our, in, in, in our right triangle along with one of the sides. And once we have the angle and one of the sides, we can use a trigonometric ratio to help us solve So let's go ahead and do that. So let's say that we let the B equal to the diameter off the sun. Yeah. And let's let R be equal to the radius of the sun. So first, let's solve for R and then substitute to find the value of D. Now, what do we already know in solving for R? Let's take one of our right triangles. So in our right, we know that we have the angle took half of 21 minutes, we also know that we have one of the sides X and the notice that X is adjacent to our angle. So X uh we can let X so since X is the adjacent side, OK. And we're solving for a side R which in our diagram is opposite to our angle. So R is the opposite side, we can solve for R using a ratio that connects or, or that links or that has the adjacent and opposite side. Now, what is that ratio? That would be the tangent? So we can solve a R using the tangent ratio. So recall that the tangent of an angle equals the opposite side divided by the adjacent side. So in this case, the tangent of our angle which is half of minutes equals the opposite side, which is R divided by the adjacent side, which is the distance X. Therefore, our side R would be equal to X multiplied by the tangent of a half of 21 minutes know that we have a value for R because we already know what the value of X is, we can solve for D because no, that tells us that or diameter D would be, which is equal to two multiplied by R would be equal to two multiplied by X multiplied by the tangent of a half of 21 minutes. So since X that distance is 141 million, 610,500 miles, we will be multiplying it by the tangent of minutes divided by two. Now, when we put this value into our calculator, then we get the diameter of the sun to be equal to 865,052 miles 0.0081. But now remember we're estimating the, the, the length of the diameter. We're estimating what it is. And in our answer choices, notice that they are all estimated to the nearest 1000 miles. So 865,052 0. miles to the nearest 1000 miles would have been equal to 865,000. Therefore, the diameter of the sun is approximately 865,000 miles. When you look back on our answer choices, that would have been answer choice. A thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Angle Measurement

Trigonometric Functions

Small Angle Approximation

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