Exercises — Trigonometric ratios in right triangle — sin, cos, tan, cosec, sec, cot

Solution 1.1 WHAT we do: just apply the definitions from SOH-CAH-TOA. WHY: each ratio is a fixed comparison of two named sides — nothing to compute beyond reading. $\sin \theta =\frac{\text{opp}}{\text{hyp}}=\frac{8}{17},\cos \theta =\frac{\text{adj}}{\text{hyp}}=\frac{15}{17},\tan \theta =\frac{\text{opp}}{\text{adj}}=\frac{8}{15}$ WHAT IT LOOKS LIKE: on Figure s01, $\sin$ is (short vertical leg)/(long slanted side), $\cos$ is (bottom leg)/(long slanted side), $\tan$ is (vertical)/(bottom).

Solution 1.2 WHAT we do: swap which leg is "opposite." WHY: opposite/adjacent are named relative to the angle. For $\varphi$, the side of length $15$ is now opposite and the side of length $8$ is now adjacent. The hypotenuse ($17$) never changes — it always faces the right angle. $\sin \varphi =\frac{15}{17},\cos \varphi =\frac{8}{17}$ Notice $\sin \varphi =\cos \theta$: the two acute angles trade sine and cosine, because $\theta$'s opposite is $\varphi$'s adjacent.

Solution 2.1 WHAT: write the definition and cross-multiply. WHY $\cos$? because $\cos$ is the only ratio linking adjacent to hypotenuse — exactly the two quantities in play. $\cos \theta =\frac{\text{adj}}{\text{hyp}}=\frac{4}{5}⟹\frac{\text{adj}}{20}=\frac{4}{5}$ $\text{adj}=20\cdot \frac{4}{5}=16\text{ cm}$

Solution 2.2 WHAT: ratios first, then Pythagoras theorem for the missing side. $\tan \theta =\frac{\text{opp}}{\text{adj}}=\frac{6}{6}=1$ A tangent of $1$ means opposite = adjacent, so $\theta =45^{\circ }$ — the perfectly balanced triangle. WHY Pythagoras: it is the one law linking all three sides, letting us fill in the hypotenuse from the two legs. ${\text{hyp}}^2=6^2+6^2=72⟹\text{hyp}=\sqrt{72}=6\sqrt{2}$ $\sin \theta =\frac{6}{6\sqrt{2}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\approx 0.707$

Solution 2.3 WHAT: treat opposite $=5$, adjacent $=12$ as a representative triangle. WHY we may: by Similar triangles, any triangle with $\tan \theta =5/12$ has the same sine and cosine — the ratio fixes the shape. $\text{hyp}=\sqrt{5^2+12^2}=\sqrt{169}=13$ $\sin \theta =\frac{5}{13},\cos \theta =\frac{12}{13}$

Solution 3.1 WHAT we know: hypotenuse (ladder) $=10$, adjacent (ground distance) $=6$. WHICH tool? adjacent + hypotenuse points straight at $\cos$. $\cos \theta =\frac{6}{10}=0.6⟹\theta =\arccos (0.6)\approx 53.13^{\circ }$ Height (opposite side) via Pythagoras: $\text{opp}=\sqrt{10^2-6^2}=\sqrt{64}=8\text{ m}$ Check: $\sin \theta =8/10=0.8$, and $0.6^2+0.8^2=1$. ✓ (This is the classic 6–8–10 triangle, a scaled 3–4–5.)

Solution 3.2 WHAT: first undo the reciprocal. WHY: $\sec$ is $1/\cos$, so recovering $\cos$ is the natural first move. $\cos \theta =\frac{1}{\sec \theta }=\frac{7}{25}=\frac{\text{adj}}{\text{hyp}}$ So adjacent $=7$, hypotenuse $=25$. Find opposite: $\text{opp}=\sqrt{25^2-7^2}=\sqrt{625-49}=\sqrt{576}=24$ $\sin \theta =\frac{24}{25},\cot \theta =\frac{\text{adj}}{\text{opp}}=\frac{7}{24}$

Solution 4.1 Step 1 — build the triangle. Opposite $=20$, hypotenuse $=29$. $\text{adj}=\sqrt{29^2-20^2}=\sqrt{841-400}=\sqrt{441}=21⟹\cos \theta =\frac{21}{29}$ Step 2 — check the identity (a built-in consistency law, from Trigonometric identities): ${\left(\frac{20}{29}\right)}^2+{\left(\frac{21}{29}\right)}^2=\frac{400+441}{841}=\frac{841}{841}=1✓$ Step 3 — $\tan \theta$ from sides: $\frac{\text{opp}}{\text{adj}}=\frac{20}{21}$. Step 4 — $\tan \theta$ from quotient identity: $\frac{\sin \theta }{\cos \theta }=\frac{20/29}{21/29}=\frac{20}{29}\cdot \frac{29}{21}=\frac{20}{21}$. Both agree. ✓ WHY they must match: the quotient identity is just the side-ratio with the shared hypotenuse ($29$) cancelling — same triangle, two viewpoints.

Solution 4.2 WHAT: replace $\csc$ and $\sec$ with their reciprocal meanings. WHY: $\csc =1/\sin$, so dividing by $\csc$ is the same as multiplying by $\sin$. $\frac{\sin \theta }{\csc \theta }=\frac{\sin \theta }{1/\sin \theta }=\sin \theta \cdot \sin \theta ={\sin }^2\theta$ $\frac{\cos \theta }{\sec \theta }={\cos }^2\theta$ $\therefore {\sin }^2\theta +{\cos }^2\theta =1\text{(Pythagorean identity)}$ Verify with $\sin =20/29,\cos =21/29$: ${\left(\frac{20}{29}\right)}^2+{\left(\frac{21}{29}\right)}^2=1$. ✓

Solution 5.1 $\tan \theta \cdot \cot \theta =\frac{\text{opp}}{\text{adj}}\cdot \frac{\text{adj}}{\text{opp}}=1$ $\sin \theta \cdot \csc \theta =\frac{\text{opp}}{\text{hyp}}\cdot \frac{\text{hyp}}{\text{opp}}=1$ WHY forced: each pair is a ratio times its own reciprocal; the two side-lengths cancel exactly, leaving $1$ regardless of the triangle's size or angle.

Solution 5.2 WHAT: horizontal distance is adjacent ($40$ m); tower height is opposite. WHICH tool: opposite over adjacent = $\tan$. $\tan \theta =\frac{\text{height}}{40}=\frac{3}{4}⟹\text{height}=40\cdot \frac{3}{4}=30\text{ m}$ The 3–4 ratio scales to $30$–$40$ (a 3–4–5 triangle with hypotenuse $50$). No calculator needed — recognising the Pythagorean triple does all the work.

Solution 5.3 WHAT: for a unit vector, horizontal component $=\cos \theta$, vertical component $=\sin \theta$. WHY: on the unit circle, the hypotenuse is $1$, so $\cos =\text{adj}/1=\text{adj}$ and $\sin =\text{opp}/1=\text{opp}$ — the components are the trig ratios. Find $\sin \theta$: adjacent $=5$, hypotenuse $=13$, so opposite $=\sqrt{13^2-5^2}=\sqrt{144}=12$, giving $\sin \theta =\frac{12}{13}$. $\text{horizontal}=\frac{5}{13},\text{vertical}=\frac{12}{13}$ ${\left(\frac{5}{13}\right)}^2+{\left(\frac{12}{13}\right)}^2=\frac{25+144}{169}=\frac{169}{169}=1✓$

One-line self-test Why does $\sin \varphi =\cos \theta$ when $\theta$ and $\varphi$ are the two acute angles of a right triangle? ::: Because $\theta$'s opposite side is $\varphi$'s adjacent side (and they share the same hypotenuse), so opp/hyp for one equals adj/hyp for the other.