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3 years ago

a sailor sights the top of a 200-foot cliff at a 17 degree angle of depression. how far is the boat from the base of the cliff

Mathematics

1 answer:

Alex_Xolod [135]3 years ago

8 0

The boat is roughly 654ft away from the base of the cliff.

Step-by-step explanation:

We have the angle 17°

The side opposite the angle is 200ft.

We want to find the length of the adjacent side.

So what equation has Opposite and Adjacent in it?

Tangent. (Remember SOH-CAH-TOA)

Tan = opp/adj

Tan(theta°) = opp/adj

Rearranging,

adj = opp/tan(theta°)

adj = 200/tan(17°)

adj = 654.17ft

And to double check you answer,

° = tan^-1(opp/adj)

° = tan%-1(200/654)

° = 17°

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$\frac{1}{4} = 5 - x^2\ \Rightarrow\ x^2 = 5 - \frac{1}{4}\ \Rightarrow\ x^2 = \frac{19}{4}\ \Rightarrow \\
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plug $x=\sqrt{19}/2$ into p( q(x) ) to get answer

$p(1/4) = p\left( q\left( \frac{ \sqrt{19} }{2} \right) \right)\ \Rightarrow\ \dfrac{4 - \left( \frac{\sqrt{19} }{2}\right)^2 }{ \left( \frac{\sqrt{19} }{2}\right)^3 } \Rightarrow \\ \\ \dfrac{4 - \frac{19}{4} }{ \frac{19\sqrt{19} }{8}} \Rightarrow \dfrac{8\left(4 - \frac{19}{4}\right) }{ 8 \cdot \frac{19\sqrt{19} }{8}} \Rightarrow \dfrac{32 - 38}{19\sqrt{19}} \Rightarrow \dfrac{-6}{19\sqrt{19}} \cdot \frac{\sqrt{19}}{\sqrt{19}}\Rightarrow$

$\dfrac{-6\sqrt{19} }{19 \cdot 19} \\ \\ \Rightarrow -\dfrac{6\sqrt{19} }{361}$

$p(1/4) = -\dfrac{6\sqrt{19} }{361}$

3 0

3 years ago

a sailor sights the top of a 200-foot cliff at a 17 degree angle of depression. how far is the boat from the base of the cliff​

a sailor sights the top of a 200-foot cliff at a 17 degree angle of depression. how far is the boat from the base of the cliff