Exercises — SOH-CAH-TOA mnemonic

We have opposite and hypotenuse. Scan the three recipes:

• SOH uses opp and hyp ✓
• CAH uses adj and hyp ✗ (no adjacent given)
• TOA uses opp and adj ✗

So we use SOH: $\sin \theta =\frac{\text{opp}}{\text{hyp}}=\frac{5}{13}\approx 0.3846$ What this means: the angle's sine — its "opposite-over-hypotenuse steepness" — is $\frac{5}{13}$. We stop here because the problem only asked which ratio.

A side that touches the angle but is not the hypotenuse is the adjacent side. The recipes containing "A" are CAH ($\cos =\text{adj}/\text{hyp}$) and TOA ($\tan =\text{opp}/\text{adj}$). So the adjacent side appears in cosine and tangent, never in sine.

Have: angle $25^{\circ }$, hypotenuse $12$. Want: the height = opposite side. Opp + Hyp → SOH. Why sine and not tangent? Because tangent needs the adjacent side, which we neither know nor want. $\sin 25^{\circ }=\frac{h}{12}\Rightarrow h=12\sin 25^{\circ }$ $h=12\times 0.4226=5.07\text{ m}$ Sense check: at a gentle $25^{\circ }$ the height should be much less than the $12$ m slope — and $5.07<12$. ✓

Have: angle $18^{\circ }$, adjacent (horizontal) $80$. Want: opposite (height). Opp + Adj → TOA. $\tan 18^{\circ }=\frac{h}{80}\Rightarrow h=80\tan 18^{\circ }=80\times 0.3249=26.0\text{ m}$

Have: opposite (height) $16$, hypotenuse (wire) $20$. Want: the angle. Opp + Hyp → SOH, but the angle is unknown, so we run sine backwards with the inverse function: $\sin \theta =\frac{16}{20}=0.8\Rightarrow \theta ={\sin }^{-1}(0.8)=53.13^{\circ }$ Here ${\sin }^{-1}$ reads "the angle whose sine is $0.8$" — it is not $1/\sin$.

Adjacent first (cosine gives it directly): $\cos \theta =\frac{\text{adj}}{10}=0.6\Rightarrow \text{adj}=6$ Opposite via Pythagoras (we know two sides now): $\text{opp}=\sqrt{10^2-6^2}=\sqrt{100-36}=\sqrt{64}=8$ So the legs are $6$ and $8$ — the classic $6$-$8$-$10$ right triangle. Cross-check with sine: $\sin \theta =\frac{8}{10}=0.8$, and $0.6^2+0.8^2=0.36+0.64=1$. ✓

The $500$ m is measured along the slope → it is the hypotenuse. (a) Altitude = opposite. Opp + Hyp → SOH: $\text{gain}=500\sin 8^{\circ }=500\times 0.1392=69.6\text{ m}$ (b) Horizontal = adjacent. Adj + Hyp → CAH: $\text{horizontal}=500\cos 8^{\circ }=500\times 0.9903=495.1\text{ m}$ Sense check: a shallow $8^{\circ }$ climb means you move mostly forward and only a little up — indeed $495.1\gg 69.6$. ✓

Each sight-line is the hypotenuse of a right triangle whose adjacent side is the $30$ m ground gap. Height of B (seen from A): Adj + angle → TOA: $h_B=30\tan 22^{\circ }=30\times 0.4040=12.12\text{ m}$ Height of A (seen from B): $h_A=30\tan 27^{\circ }=30\times 0.5095=15.29\text{ m}$ Two separate triangles, same shared base — this is the seed idea behind surveying.

Split an equilateral triangle (side $2$) down the middle → a $30$-$60$-$90$ triangle with base $1$, height $\sqrt{3}$ (since $\sqrt{2^2-1^2}=\sqrt{3}$). For the $60^{\circ }$ angle at the base: $\tan 60^{\circ }=\frac{\text{opp}}{\text{adj}}=\frac{\sqrt{3}}{1}=\sqrt{3}$ This is an exact value from geometry — see special angles. Now the tower: Adj $=15$, angle $=60^{\circ }$, want opposite: $h=15\tan 60^{\circ }=15\sqrt{3}\approx 25.98\text{ m}$ Exact answer $15\sqrt{3}$ m; decimal $25.98$ m.

Both sight-lines share the same adjacent distance $x$. Let the top height be $H$ and window height $W$. $H=x\tan 50^{\circ },W=x\tan 32^{\circ }$ The gap between them is given: $H-W=18\Rightarrow x\tan 50^{\circ }-x\tan 32^{\circ }=18$ Factor out $x$ (this is the key algebra step — same $x$ in both): $x(\tan 50^{\circ }-\tan 32^{\circ })=18$ $x=\frac{18}{\tan 50^{\circ }-\tan 32^{\circ }}=\frac{18}{1.19175-0.62487}=\frac{18}{0.56688}=31.75\text{ m}$ Sense check: $H=31.75\tan 50^{\circ }\approx 37.8$ m, $W=31.75\tan 32^{\circ }\approx 19.8$ m, gap $\approx 18.0$ m. ✓

Picture the hypotenuse fixed at length $1$ and let $\theta$ shrink toward $0^{\circ }$: the triangle flattens.

• As $\theta \to 0^{\circ }$: opposite $\to 0$, adjacent $\to$ hypotenuse. So $\sin \theta \to 0$, $\cos \theta \to 1$, $\tan \theta =\frac{\text{opp}}{\text{adj}}\to 0$.
• As $\theta \to 90^{\circ }$: now the triangle is tall and thin; opposite $\to$ hypotenuse, adjacent $\to 0$. So $\sin \theta \to 1$, $\cos \theta \to 0$, and $\tan \theta =\frac{\text{opp}}{\text{adj}}\to \infty$ (adjacent shrinking to zero makes the ratio blow up — tangent has no finite value at $90^{\circ }$).

Confirm one edge value using an exact special angle, $\tan 45^{\circ }$: an isosceles right triangle has equal legs, so $\tan 45^{\circ }=\frac{\text{opp}}{\text{adj}}=1$. This sits neatly between $\tan 0^{\circ }=0$ and the blow-up at $90^{\circ }$. The full picture of this behaviour is drawn in the sine–cosine graphs and grounded in the unit circle.

The ratio using opposite and hypotenuse is sine (SOH). To find an angle from a known ratio you apply an ==inverse trig function (${\sin }^{-1}$, ${\cos }^{-1}$, ${\tan }^{-1}$)==. $\tan \theta$ is undefined at $\theta =$ ==$90^{\circ }$ because the adjacent side becomes zero (division by zero)==. Which ratio has value $\sqrt{3}$ at $60^{\circ }$? ::: $\tan 60^{\circ }$ If $\cos \theta =0.6$ and hyp $=10$, the adjacent side is ::: $6$