Question #4e9e7

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

SOH:
S = sin `theta`
O = opposite
H = hypotenuse

So, the ratio is sin `theta` = `"opposite"/"hypotenuse"`

CAH:
C = cos `theta`
A = adjacent
H = hypotenuse

The ratio is cos `theta` = `"adjacent"/"hypotenuse"`

TOA:
T = tan `theta`
O = opposite
A = adjacent

The ratio is tan `theta` = `"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)` m. Then the angle of elevation `theta` is the angle opposite the 15 m.

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

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

tan `theta` = `"opposite"/"adjacent"`
tan `theta` = `15/(15sqrt(3))`
`theta` = `tan^-1(15/(15sqrt(3)))`
`theta` = `pi/6` radians or `30` degrees