Every right-angled triangle has a hypotenuse (opposite the right angle, always the longest side). Once you pick an angle $\theta$ to work with (not the right angle), the other two sides are named relative to it:

        /|
       / |
  Hyp /  | Opp   (Opposite the angle θ)
     /   |
    /θ___|
     Adj    (Adjacent to the angle θ)

Label the two sides you know or want (relative to $\theta$), then pick whichever of SOH, CAH or TOA contains both of those letters.

Set up the ratio with the unknown side as the subject, then rearrange.

In a right-angled triangle, $\theta=35^\circ$ and the hypotenuse is $12\text{ cm}$. Find the side opposite $\theta$, to 3 s.f.:

1. Known: hypotenuse. Wanted: opposite. This pair appears in SOH.
2. $\sin(35^\circ) = \dfrac{x}{12}$.
3. $x = 12\times\sin(35^\circ) = 6.8829\ldots \approx 6.88\text{ cm}$.

Set up the ratio as normal, then undo it with the matching inverse trig function.

A triangle has opposite side $7\text{ cm}$ and adjacent side $9\text{ cm}$, relative to $\theta$. Find $\theta$ to 1 d.p.:

1. Known sides: opposite and adjacent — this pair appears in TOA.
2. $\tan\theta = \dfrac{7}{9}$.
3. $\theta = \tan^{-1}\left(\dfrac{7}{9}\right) = 37.874\ldots^\circ \approx 37.9^\circ$.

An angle of elevation is measured upward from the horizontal to a point above (e.g. looking up at the top of a tower); an angle of depression is measured downward from the horizontal to a point below (e.g. looking down from a cliff at a boat). Because the horizontal lines at the two positions are parallel, the angle of elevation from the ground and the angle of depression from the higher point are always equal (alternate angles).

Sketch the right-angled triangle formed by the two legs of the journey, work out the required angle using SOH/CAH/TOA, then convert that angle into a three-figure bearing based on the direction of travel.

A ship sails $15\text{ km}$ east, then $9\text{ km}$ north. Find the bearing of its final position from the start, to the nearest degree:

1. The east leg ($15$) and north leg ($9$) are perpendicular. The angle $\theta$ from north uses adjacent ($9$) and opposite ($15$) — TOA:

1. $\theta = \tan^{-1}\left(\dfrac{15}{9}\right) = 59.036\ldots^\circ \approx 59^\circ$.
2. Since the ship has moved into the north-east region, the bearing is:

A kite string makes an angle of $52^\circ$ with the horizontal ground, and the string is $30\text{ m}$ long. Find the height of the kite above the ground, to 3 s.f.

Solution

The string is the hypotenuse; the height is opposite the $52^\circ$ angle — this is SOH: