Question #4e9e7

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Question #4e9e7

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Answer 1 · 2017-11-06T05:57:27

tan $θ$ $theta$ = $\frac{opposite}{adjacent}$ $"opposite"/"adjacent"$

Explanation:

A good memory trick for sin, cos, and tan ratios for right angle triangles is "SOH CAH TOA".

SOH:
S = sin $θ$ $theta$
O = opposite
H = hypotenuse

So, the ratio is sin $θ$ $theta$ = $\frac{opposite}{hypotenuse}$ $"opposite"/"hypotenuse"$

CAH:
C = cos $θ$ $theta$
A = adjacent
H = hypotenuse

The ratio is cos $θ$ $theta$ = $\frac{adjacent}{hypotenuse}$ $"adjacent"/"hypotenuse"$

TOA:
T = tan $θ$ $theta$
O = opposite
A = adjacent

The ratio is tan $θ$ $theta$ = $\frac{opposite}{adjacent}$ $"opposite"/"adjacent"$

For this particular problem, draw yourself a right angle triangle with the vertical side (height of the tree) being 15 m and the base of the triangle (the shadow) being 15 $\sqrt{3}$ $sqrt(3)$ m. Then the angle of elevation $θ$ $theta$ is the angle opposite the 15 m.

The two known sides are opposite (15m) and adjacent (15 $\sqrt{3}$ $sqrt(3)$ m) to the unknown angle of elevation $θ$ $theta$ . Looking back at the ratios defined earlier, we see that only tan $θ$ $theta$ involves both opposite and adjacent sides.

So, to solve the problem, we simply substitute in our values and solve for $θ$ $theta$ :

tan $θ$ $theta$ = $\frac{opposite}{adjacent}$ $"opposite"/"adjacent"$
tan $θ$ $theta$ = $\frac{15}{15 \sqrt{3}}$ $15/(15sqrt(3))$
$θ$ $theta$ = ${tan}^{- 1} (\frac{15}{15 \sqrt{3}})$ $tan^-1(15/(15sqrt(3)))$
$θ$ $theta$ = $\frac{π}{6}$ $pi/6$ radians or $30$ $30$ degrees

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Question #4e9e7

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