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Lynn is trying to determine how far away Student B is from the balloon. He decides to use the

equation shown below. Is his equation correct? Why or why not?

5

cos 60º =

BIV x

User Acs
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4.5k points

2 Answers

2 votes

Answer:

Did you get the answer If so please give it to me.

Explanation:

User Cutiko
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4.6k points
5 votes

Answer:

See Explanation

Explanation:

The question is incomplete as the image that illustrates the scenario is not given.

However, I can deduce that the question is about a right-angled triangle.

So, I will give a general explanation on how to find each of the side of the triangle, given a side and an angle.

For triangle A (solve for b)

Using cosine formula.


\cos \theta = (Adjacent)/(Hypotenuse)


\cos 60= (5)/(b)

Make b the subject


b= (5)/(\cos 60)

For triangle B (solve for b)

Using cosine formula.


\sin \theta = (Opposite)/(Hypotenuse)


\sin 60= (b)/(5)

Make b the subject


b = 5\sin 60

For triangle C (solve for b)

Using cosine formula.


\tan \theta = (Opposite)/(Adjacent)


\tan 60= (b)/(5)

Make b the subject


b = 5\tan 60

Lynn is trying to determine how far away Student B is from the balloon. He decides-example-1
User Aeskreis
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