Trigonometry, Heights & Distances

The six ratios and their reciprocal and quotient relations; the standard-angle table; the three Pythagorean identities; complementary-angle results; signs in the four quadrants and values beyond 90°; simplifying expressions and proving identities; maximum and minimum of common expressions; angles of elevation and depression, and the standard heights-and-distances patterns — the shadow, the two-position tower, the building and the observer's height, the moving object — with fully worked examples.

Trigonometry, Heights & Distances

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📑 Contents (5 sections)

Last reviewed 22 Sept 2026 · 5 min read

The six ratios

In a right-angled triangle, with respect to an acute angle $θ$ :

sin θ = \frac{opposite}{hypotenuse}, cos θ = \frac{adjacent}{hypotenuse}, tan θ = \frac{opposite}{adjacent}

with $csc$ , $sec$ and $cot$ as the reciprocals of $sin$ , $cos$ and $tan$ . Also $tan θ = \frac{sin θ}{cos θ}$ and $cot θ = \frac{cos θ}{sin θ}$ .

Standard angles

$θ$	$0°$	$30°$	$45°$	$60°$	$90°$
$sin θ$	0	$\frac{1}{2}$	$\frac{1}{2}$	$\frac{3}{2}$	1
$cos θ$	1	$\frac{3}{2}$	$\frac{1}{2}$	$\frac{1}{2}$	0
$tan θ$	0	$\frac{1}{3}$	1	$3$	undefined

Identities

FormulaThe three that matter

sin^{2} θ + cos^{2} θ = 1, 1 + tan^{2} θ = sec^{2} θ, 1 + cot^{2} θ = csc^{2} θ

Useful rearrangements: $sec^{2} θ - tan^{2} θ = 1$ and $csc^{2} θ - cot^{2} θ = 1$ , so

sec θ - tan θ = \frac{1}{sec θ + tan θ}

FormulaComplementary angles

sin (90° - θ) = cos θ, tan (90° - θ) = cot θ, sec (90° - θ) = csc θ

So $sin^{2} 20° + sin^{2} 70° = 1$ and $tan 1° tan 2° \dots tan 89° = 1$ , because the terms pair off.

Signs and larger angles

All ratios are positive in the first quadrant; only $sin$ and $csc$ in the second, only $tan$ and $cot$ in the third, only $cos$ and $sec$ in the fourth ("add sugar to coffee"). For values past $90°$ : $sin (180° - θ) = sin θ$ and $cos (180° - θ) = - cos θ$ .

Maximum and minimum

$- 1 \leq sin θ \leq 1$ and $- 1 \leq cos θ \leq 1$
$a sin θ + b cos θ$ lies between $- a^{2} + b^{2}$ and $a^{2} + b^{2}$
$sin θ cos θ = \frac{1}{2} sin 2 θ$ , so its maximum is $\frac{1}{2}$
$sec^{2} θ + csc^{2} θ \geq 4$ , with equality at $45°$

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