Question

Point A is 20 ft from a tree and point B, on the opposite side of the tree, is 10 ft from that tree, as shown in the figure. The distance from point A to the top of a tree is 29 ft. The angle of elevation to the top of the tree from point A is 47°.

To the nearest foot, how far is the top of the tree from point B?

To the nearest degree, what is the angle of elevation to the top of the tree from point B?

Answer 1

Answer:

The answer is 23.544ft and 64.3 degrees. ( I know that they don't match with any options but I think I'm using the correct formulas and this should be the right answer )

Step-by-step explanation:

• Let the top of the tree be point C.
• Firstly, we need to use cosine rule. The cosine rule is as follows:
• ${cb}^{2} = {(ab)}^{2} + {(ac)}^{2} - 2(ab)(ac) \cos(cab)$
• AB is 30ft and AC is 29ft. After inserting everything in the formula and putting it in the calculator we get the length of CB as 23.544ft.
• After finding the length we can find the angle by using the sine rule. The sine rule is as follows:
• $\frac{ac}{ \sin(abc) } = \frac{cb}{ \sin(bac) }$
• After using the formula above the inserted values will be:
• $\frac{29}{ \sin(abc) } = \frac{23.544}{ \sin(47) }$
• Now using this we get the angle and that is 64.268 rounded off to 64.3degrees.