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AlladinOne [14]
2 years ago
8

The angle of elevation to a nearby tree from a point on the ground is measured to be 66^{\circ} ∘ . How tall is the tree if the

point on the ground is 68 feet from the tree? Round your answer to the nearest tenth of a foot if necessary.
Mathematics
1 answer:
Nimfa-mama [501]2 years ago
6 0

By comparing with a right triangle, we will see that the height of the tree is 62.12ft

<h3>How tall is the tree?</h3>

We can compare the situation with a right triangle. Such that the hypotenuse measures 68 ft, and an angle measures 66°.

We want to get the height of the tree, which would be the opposite cathetus, then we can use the relation:

Sin(a) = (opposite cathetus)/hypotenuse.

Sin(66°) = H/68ft

Tan(66°)*68ft = H = 62.12ft

The height of the tree is 62.12ft

If you want to learn more about right triangles, you can read:

brainly.com/question/2217700

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~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\dotfill &\£95000\\ P=\textit{original amount deposited}\dotfill &\£80000\\ r=rate\to r\%\to \frac{r}{100}\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{yearly, thus once} \end{array}\dotfill &1\\ t=years\dotfill &5 \end{cases}

95000=80000\left(1+\frac{~~ \frac{r}{100}~~}{1}\right)^{1\cdot 5}\implies \cfrac{95000}{80000}=\left( 1+\cfrac{r}{100} \right)^5 \\\\\\ \cfrac{19}{16}=\left( 1+\cfrac{r}{100} \right)^5\implies \sqrt[5]{\cfrac{19}{16}}=1+\cfrac{r}{100}\implies \sqrt[5]{\cfrac{19}{16}}=\cfrac{100+r}{100} \\\\\\ 100\sqrt[5]{\cfrac{19}{16}}=100+r\implies 100\sqrt[5]{\cfrac{19}{16}}-100=r\implies 3.5\approx r

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