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Nikolay [14]
3 years ago
15

An airplane descends 2.2 miles to an elevation of 8.55 miles. Find the elevation of the plane before its descent.

Mathematics
1 answer:
lutik1710 [3]3 years ago
7 0

Answer:

10.75 miles was the elavation before it's decsecent.

Step-by-step explanation:

Hope this helps! :)

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Find the exact value of each trigonometric function for the given angle θ.
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Answer:

\sin (240^\circ)=-\dfrac{\sqrt{3}}{2},\cos (240^\circ)=-\dfrac{1}{2},\tan (240^\circ)=\sqrt{3},\cot (240^\circ)=\dfrac{1}{\sqrt{3}},\sec (240^\circ)=-2,\csc (240^\circ)=\dfrac{2}{\sqrt{3}}.

Step-by-step explanation:

The given angle is 240 degrees.

We need to find the exact value of each trigonometric function for the given angle θ.

Since \theta=240, it means θ lies in 3rd quadrant. In 3d quadrant only tan and cot are positive.

\sin (240^\circ)=\sin (180^\circ+60^\circ)=-\sin (60^\circ)=-\dfrac{\sqrt{3}}{2}

\cos (240^\circ)=\cos (180^\circ+60^\circ)=-\cos (60^\circ)=-\dfrac{1}{2}

\tan (240^\circ)=\tan (180^\circ+60^\circ)=\tan (60^\circ)=\sqrt{3}

\cot (240^\circ)=\cot (180^\circ+60^\circ)=\cot (60^\circ)=\dfrac{1}{\sqrt{3}}

\sec (240^\circ)=\sec (180^\circ+60^\circ)=-\sec (60^\circ)=-2

\csc (240^\circ)=\csc (180^\circ+60^\circ)=-\csc (60^\circ)=-\dfrac{2}{\sqrt{3}}

Therefore, \sin (240^\circ)=-\dfrac{\sqrt{3}}{2},\cos (240^\circ)=-\dfrac{1}{2},\tan (240^\circ)=\sqrt{3},\cot (240^\circ)=\dfrac{1}{\sqrt{3}},\sec (240^\circ)=-2,\csc (240^\circ)=-\dfrac{2}{\sqrt{3}}.

8 0
3 years ago
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