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

Before anything, one reminder of the picture every example leans on.

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The scenario matrix

Cell	Scenario class	What's given	What's asked	Example
A	Two legs known	$o$, $a$	all six ratios	Ex 1
B	One ratio + one side	$\sin \theta$, $h$	missing sides	Ex 2
C	Reciprocal ratio given	$\sec \theta$	$\sin ,\cos ,\tan$	Ex 3
D	Ratio → other ratio (no triangle)	$\tan \theta$	$\sin ,\cos$	Ex 4
E	Limiting / degenerate	$\theta \to 0^{\circ }$, $\theta \to 90^{\circ }$	behaviour of ratios	Ex 5
F	Special angles (exact values)	$\theta =45^{\circ }$, $30^{\circ }$	exact ratios	Ex 6
G	Real-world word problem	ladder height	angle & sides	Ex 7
H	Exam twist (mixed reciprocals)	$\csc \theta +\cot \theta$	isolate $\cos \theta$	Ex 8

Eight examples, eight cells — no gaps.

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Cell A — both legs known

Forecast: guess — will the hypotenuse be a whole number here? Jot a yes/no before reading on.

1. Find $h$ with Pythagoras theorem. $h^2=o^2+a^2=8^2+15^2=64+225=289⟹h=17$ Why this step? The three ratios that use $h$ (sin, cos, and their reciprocals) can't be written until we know $h$. Pythagoras is the only tool that gets a side from two other sides in a right triangle.

2. Write the primary three. $\sin \theta =\frac{o}{h}=\frac{8}{17},\cos \theta =\frac{a}{h}=\frac{15}{17},\tan \theta =\frac{o}{a}=\frac{8}{15}$ Why this step? These are the definitions from SOH-CAH-TOA — read straight off the labelled triangle.

3. Flip each for the reciprocals. $\csc \theta =\frac{17}{8},\sec \theta =\frac{17}{15},\cot \theta =\frac{15}{8}$ Why this step? Cosec/sec/cot are literally sin/cos/tan turned upside down — no new geometry needed.

Verify: the Pythagorean identity must hold: ${\sin }^2\theta +{\cos }^2\theta =\frac{64}{289}+\frac{225}{289}=\frac{289}{289}=1✓$ (Your forecast: yes, $h=17$ was whole — an $8,15,17$ triple.)

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Cell B — one ratio and one side

Forecast: which side does $\cos$ hand you directly — opposite or adjacent?

1. Use the definition to get the adjacent side. $\cos \theta =\frac{a}{h}=\frac{9}{41}⟹\frac{a}{82}=\frac{9}{41}⟹a=82\cdot \frac{9}{41}=18\text{ cm}$ Why this step? $\cos =\text{adjacent}/\text{hypotenuse}$, and we know the hypotenuse, so cross-multiplying frees $a$.

2. Get the opposite side with Pythagoras. $o^2=h^2-a^2=82^2-18^2=6724-324=6400⟹o=80\text{ cm}$ Why this step? We now know two sides; Pythagoras is again the bridge to the third.

Verify: ${\sin }^2\theta +{\cos }^2\theta ={\left(\frac{80}{82}\right)}^2+{\left(\frac{9}{41}\right)}^2=\frac{6400}{6724}+\frac{324}{6724}=\frac{6724}{6724}=1✓$ (units cm consistent).

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Cell C — a reciprocal ratio is given

Forecast: secant flips cosine. So which ratio drops out first with zero effort?

1. Undo the reciprocal to get cosine. $\sec \theta =\frac{1}{\cos \theta }=\frac{25}{7}⟹\cos \theta =\frac{7}{25}$ Why this step? $\sec$ is defined as $1/\cos$; flipping a fraction is the fastest legal move.

2. Read cosine as sides, then find the missing leg. $\cos \theta =\frac{a}{h}=\frac{7}{25}$ means $a=7$, $h=25$ (up to scale). Then $o=\sqrt{h^2-a^2}=\sqrt{625-49}=\sqrt{576}=24$ Why this step? Reconstructing the triangle turns one ratio into all three sides — Pythagoras supplies the leg we lack.

3. Read off the remaining ratios. $\sin \theta =\frac{o}{h}=\frac{24}{25},\tan \theta =\frac{o}{a}=\frac{24}{7}$ Why this step? With all three sides in hand every ratio is a direct definition.

Verify: ${\sin }^2\theta +{\cos }^2\theta =\frac{576}{625}+\frac{49}{625}=\frac{625}{625}=1✓$

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Cell D — ratio to ratio, no triangle drawn

Forecast: we have one equation. What second equation lets us pin down two unknowns?

1. Bring in the identity that links tan to sin/cos. $\tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{3}{4}⟹\sin \theta =\frac{3}{4}\cos \theta$ Why this step? The quotient identity converts the given $\tan$ into a relation between $\sin$ and $\cos$ — pure algebra, no geometry.

2. Feed that into the Pythagorean identity. ${\sin }^2\theta +{\cos }^2\theta =1⟹{\left(\frac{3}{4}\cos \theta \right)}^2+{\cos }^2\theta =1$ $\frac{9}{16}{\cos }^2\theta +{\cos }^2\theta =\frac{25}{16}{\cos }^2\theta =1⟹{\cos }^2\theta =\frac{16}{25}$ Why this step? Two equations, two unknowns — substitution collapses them into one solvable equation.

3. Take the positive root (acute angle) and back-substitute. $\cos \theta =\frac{4}{5},\sin \theta =\frac{3}{4}\cdot \frac{4}{5}=\frac{3}{5}$ Why this step? For an acute $\theta$ both sin and cos are positive, so we keep $+$; step 1 then gives sin.

Verify: $\frac{\sin \theta }{\cos \theta }=\frac{3/5}{4/5}=\frac{3}{4}=\tan \theta ✓$

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Cell E — limiting and degenerate cases

Forecast: a right triangle can't actually reach $0^{\circ }$ or $90^{\circ }$ (it would flatten). So we ask what the ratios approach.

1. Squash the triangle: $\theta \to 0^{\circ }$. The opposite side shrinks to nothing while adjacent $\approx$ hypotenuse. So $\sin \theta =\frac{o}{h}\to 0,\cos \theta =\frac{a}{h}\to 1,\tan \theta =\frac{o}{a}\to 0$ Why this step? Reading the definitions off the flattening picture tells us the limits directly.

2. Spot the blow-ups at $\theta \to 0^{\circ }$. Since $o\to 0$, any ratio dividing by $o$ explodes: $\csc \theta =\frac{h}{o}\to \infty ,\cot \theta =\frac{a}{o}\to \infty$ while $\sec \theta =\frac{h}{a}\to 1$. Why this step? Dividing a fixed length by a vanishing length grows without bound — that's why csc and cot "go to infinity."

3. Tip it the other way: $\theta \to 90^{\circ }$. Now adjacent $\to 0$, opposite $\approx$ hypotenuse: $\sin \theta \to 1,\cos \theta \to 0,\tan \theta =\frac{o}{a}\to \infty ,\sec \theta =\frac{h}{a}\to \infty$ Why this step? By the mirror-image logic, dividing by the vanishing adjacent makes tan and sec blow up.

Verify (sanity, at $\theta =89^{\circ }$): $\sin 89^{\circ }\approx 0.9998$ (near $1$) and $\tan 89^{\circ }\approx 57.29$ (large) — matches the "blow-up" claim. We check the exact limit values numerically in VERIFY.

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Cell F — special (exact) angles

Forecast: at $45^{\circ }$ the triangle is symmetric — what does that force the two legs to be?

1. The $45^{\circ }$ case: an isosceles right triangle. Equal acute angles $\Rightarrow$ equal legs, say both $=1$. Then $h=\sqrt{1^2+1^2}=\sqrt{2}$. $\sin 45^{\circ }=\cos 45^{\circ }=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2},\tan 45^{\circ }=\frac{1}{1}=1$ Why this step? Symmetry gives the leg lengths for free; Pythagoras gives $h$; then read definitions.

2. The $30^{\circ }$ case: half of an equilateral triangle. Split an equilateral triangle of side $2$ down the middle. The half has hypotenuse $2$, the short (opposite to $30^{\circ }$) side $1$, and the long side $\sqrt{2^2-1^2}=\sqrt{3}$. $\sin 30^{\circ }=\frac{1}{2},\cos 30^{\circ }=\frac{\sqrt{3}}{2},\tan 30^{\circ }=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$ Why this step? Cutting a known equilateral triangle produces a right triangle with exact side lengths, no measuring.

Verify: ${\sin }^{2}45^{\circ }+{\cos }^{2}45^{\circ }=\frac{1}{2}+\frac{1}{2}=1$; and ${\sin }^{2}30^{\circ }+{\cos }^{2}30^{\circ }=\frac{1}{4}+\frac{3}{4}=1✓$

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Cell G — real-world word problem

Forecast: where is the right angle hiding in this picture?

1. Model it as a right triangle. The wall meets the ground at $90^{\circ }$. Ladder = hypotenuse $=10$, ground distance = adjacent $=6$, wall height = opposite $=H$. (This is exactly angle of elevation geometry.) Why this step? Naming physical parts as $h$, $a$, $o$ lets us reuse every ratio we know.

2. Height by Pythagoras. $H=\sqrt{h^2-a^2}=\sqrt{10^2-6^2}=\sqrt{100-36}=\sqrt{64}=8\text{ m}$ Why this step? Two sides known, third wanted — Pythagoras again.

3. Angle from a ratio. $\cos \theta =\frac{a}{h}=\frac{6}{10}=0.6⟹\theta =\arccos (0.6)\approx 53.13^{\circ }$ Why $\arccos$? We know the cosine and want the angle — $\arccos$ answers "which angle has this cosine?", undoing cos. (See Solving right triangles.)

Verify: check with sine of the found angle: $\sin \theta =\frac{H}{h}=\frac{8}{10}=0.8$, and $0.6^2+0.8^2=1✓$; units are metres throughout.

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Cell H — exam twist with mixed reciprocals

Forecast: both terms share the same denominator — what is it?

1. Rewrite everything over $\sin \theta$. $\csc \theta +\cot \theta =\frac{1}{\sin \theta }+\frac{\cos \theta }{\sin \theta }=\frac{1+\cos \theta }{\sin \theta }=3$ Why this step? Reciprocal and quotient definitions turn a scary mix into one clean fraction in sin and cos.

2. Clear the fraction, then use the Pythagorean identity. $1+\cos \theta =3\sin \theta ⟹(1+\cos \theta {)}^2=9{\sin }^2\theta =9(1-{\cos }^2\theta )$ Why this step? We have two unknowns; replacing ${\sin }^2\theta$ by $1-{\cos }^2\theta$ leaves an equation in $\cos \theta$ only.

3. Solve the resulting equation. Let $c=\cos \theta$. Note $9(1-c^2)=9(1-c)(1+c)$: $(1+c{)}^2=9(1-c)(1+c)⟹1+c=9(1-c)⟹10c=8⟹c=\frac{4}{5}$ Why this step? Factoring lets us cancel the common $(1+c)$ (nonzero for acute $\theta$), collapsing to a linear equation.

Verify: with $\cos \theta =\frac{4}{5}$, an acute $\theta$ gives $\sin \theta =\frac{3}{5}$; then $\csc \theta +\cot \theta =\frac{5}{3}+\frac{4/5}{3/5}=\frac{5}{3}+\frac{4}{3}=\frac{9}{3}=3✓$

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Recall Quick self-test across the matrix

Which ratios blow up as $\theta \to 0^{\circ }$? ::: $\csc \theta$ and $\cot \theta$ (they divide by the vanishing opposite side). From $\sec \theta =25/7$, what is $\cos \theta$? ::: $7/25$ (secant is the reciprocal of cosine). Exact value of $\tan 45^{\circ }$? ::: $1$ (equal legs). In the ladder problem, which side is the hypotenuse? ::: the ladder itself. Why can $\tan 90^{\circ }$ not be a number? ::: adjacent is $0$, so $o/a$ divides by zero — undefined.

Related build-outs: Similar triangles (why ratios ignore size), Unit circle definition of trig functions (how these extend past $90^{\circ }$), Components of vectors (sin/cos as projections).