Question

A surveyor is measuring distances in a level yard near a road. A fence post is due north from the surveyor and is 18.5 m away from him. A street sign is to the right of the fence post from the surveyor's view, and is a distance of 23.9 m from him. The surveyor measures the angle between the fence post and the street sign to be 22.4°. (Due to the nature of this problem, do not use rounded intermediate values in your calculations—including answers submitted in WebAssign.) (a) What is the distance (in m) between the fence post and the street sign? m (b) What is the angle (in degrees) between the surveyor and the street sign, as measured from the position of the fence post? Hint: Draw a top-down view of the situation. ° (c) What is the angle (in degrees) between the surveyor and the fence post, as measured from the position of the street sign? ° (d) An astronomer measures the angle between two stars in the sky to be 22.4° apart. The distance from Earth to Star A is 18.5 ly. The distance from Earth to star B, which appears to the right of star A, is 23.9 ly. Imagine an alien living on a planet orbiting star A. From the alien's perspective, what is the angle (in degrees) in the sky between star B and our Sun?

Answer 1

Answer:

a) d=9.79m

b) ∠=68.48°

c) ∠ =89.12°

d) ∠ =68.48°

Explanation:

So, in order to solve this problem, we need to start by drawing a sketch of what the problem looks like. (see attached picture)

a)

In order to find the distance between the fence and the street sign, we must use the law of cosines, which looks like this:

$A=\sqrt{B^{2}+C^{2}-2BCcos(\theta)}$

depending on how the sides of the triangle are named.

so when plugging the sides of the triangle in, we get that:

$A=\sqrt{(18.5m)^{2}+(23.9m)^{2}-2(18.5m)(23.9)cos(22.4^{o})}$

Which yields:

$A= 9.79 m$

b)  

For part b we can make use of law of sines to find the angle the problem is asking about, so we get:

$\frac{sin \theta}{23.9}=\frac{sin 22.4^{o}}{9.79m}$

When solving for  we get:

$\theta=sin^{-1}(\frac{23.9}{9.79}sin(22.4^{o}))$

which yields:

$\theta = 68.48^{o}$

c)

Since we are talking about a triangle, for part c, all I need to do is subtract the angles I know from 180° and that will give me the missing angle:

∠ = 180°-68.48°-22.4° = 89.12°

d)

Finally, if we sketched the situation presented by part d of the problem (see attached picture) we will be able to see that it's basically the same problem, all it did was change the story but basically it's the same. So we follow the same procedure as in parts a and b to get the answer of 68.48°.