Can somebody help with these three questions plz

Mathematics

1 answer:

liraira [26]3 years ago

7 0

$\bf \sqrt{2}\left(3\sqrt{2}+\sqrt{18} \right)~~
\begin{cases}
18=2\cdot 3\cdot 3\\
\qquad 2\cdot 3^2
\end{cases}\implies \sqrt{2}\left(3\sqrt{2}+\sqrt{2\cdot 3^2} \right)
\\\\\\
\sqrt{2}\left(3\sqrt{2}+3\sqrt{2} \right)\implies \sqrt{2}\left(6\sqrt{2} \right)\implies 6\left(\sqrt{2}\right)^2\implies 6\cdot 2\implies \stackrel{rational}{12}$

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Kim is watching a space shuttle launch from an observation spot 2 miles away. Find the angle of elevation from Kim to the space

hammer [34]

Answer:

Step-by-step explanation:

Find the diagrammatic representation in the attachment given.

We can see that the diagram is a right angled triangle. To get the angle of elevation, we will use one of the trigonometry identity SOH, CAH, TOA.

Given the adjacent of the triangle to be 2miles (observation spot from the point of launch)and the opposite (side facing the elevation angle) as 2.1miles (the height), we will use TOA.

tan∅ = opposite/adjacent

tan∅ = 2.1/2

tan∅ = 1.05

∅ = tan⁻¹1.05

∅ = 46.4°

<em>Hence the angle of elevation from Kim to the space shuttle, which is at a height of 2.1 miles is approximately 46.4°</em>