Solving a right-angled triangle means calculating its main elements: the acute angles ($\hat{B}$ and $\hat{C}$) and the sides ($a$, $b$, $c$). To do this, it is necessary to provide two information about the triangle, at least one of which is the measurement of a segment connected to it (side, height, median, angle bisector etc.) — only with angles, we know the shape but not the size.

The story

Solving right triangles was, for centuries, the daily work of surveyors, navigators and artillerymen. To measure angles on the ground, the Englishman Leonard Digges described an instrument he called theodelitus — the ancestor of the theodolite, which to this day appears on the streets on a tripod. The description appeared in Pantometria (“Universal Measurement”), a book that Leonard left unfinished and that his son, Thomas Digges, completed and published in 1571. With a measured angle and a known distance, the trigonometric tables did the rest.

Measuring a tower without climbing it: with the theodolite you measure the angle $\alpha$ from the sight to the top and with the chain, the distance $d$ to the base; the height above the instrument is $h = d\operatorname{tg}\alpha$.

Sources: Thomas Digges (MacTutor) · L. Digges and T. Digges, Pantometry (1571).

In everyday life

  • Firefighter ladder. A 20 m ladder inclined at $70^\circ$ reaches $20\operatorname{sen} 70^\circ \approx 18{,}8$ m in height (plus the height of the truck).
  • Access ramp. To overcome 80 cm with 8,33% slope, the ramp needs $0{,}80 \div 0{,}0833 \approx 9{,}6$ m of horizontal projection.
  • Kites and drones. With 100 m of line at $35^\circ$ from the ground, a kite is about $100\operatorname{sen} 35^\circ \approx 57$ m high (the actual line curves, so this is an overhead estimate).
The firefighters’ ladder: with 20 m to $70^\circ$, the tip reaches $20\operatorname{sen}70^\circ \approx 18{,}8$ m.

Theory

There are five classic problems for solving right-angled triangles. In all, $\hat{A} = 90^\circ$.

Exemplo1st problem — given the hypotenuse a and an acute angle B

$$\hat{C} = 90^\circ - \hat{B} \qquad b = a\operatorname{sen}\hat{B} \qquad c = a\cos\hat{B}$$

Exemplo2nd problem — given the hypotenuse a and a side b

$$c = \sqrt{a^2 - b^2}$$$$\operatorname{sen}\hat{B} = \frac{b}{a} \Rightarrow \hat{B} = \operatorname{arc\,sen}\frac{b}{a}$$$$\cos\hat{C} = \frac{b}{a} \Rightarrow \hat{C} = \operatorname{arc\,cos}\frac{b}{a}$$

Exemplo3rd problem — given one side b and the adjacent angle C

$$\hat{B} = 90^\circ - \hat{C} \qquad c = b \cdot \operatorname{tg}\hat{C} \qquad a = \frac{b}{\cos\hat{C}}$$

Exemplo4th problem — given one side b and the opposite angle B

$$\hat{C} = 90^\circ - \hat{B} \qquad c = \frac{b}{\operatorname{tg}\hat{B}} \qquad a = \frac{b}{\operatorname{sen}\hat{B}}$$

Exemplo5th problem — given the two legs b and c

$$a = \sqrt{b^2 + c^2}$$$$\operatorname{tg}\hat{B} = \frac{b}{c} \Rightarrow \hat{B} = \operatorname{arc\,tg}\frac{b}{c}$$$$\operatorname{tg}\hat{C} = \frac{c}{b} \Rightarrow \hat{C} = \operatorname{arc\,tg}\frac{c}{b}$$

AtençãoDice that are not sides

When the data are other segments — height, angle bisector, perimeter, radius of the inscribed circle —, first set up the equations that connect them to the sides (the metric relations in the first section help a lot) and only then fall back on one of the five problems. This is what happens in exercises 352 to 359.

ExemploA complete example

Solve the right triangle with $a = 10$ and $\hat{B} = 30^\circ$ (1st problem): $\hat{C} = 60^\circ$, $b = 10\operatorname{sen} 30^\circ = 5$ and $c = 10\cos 30^\circ = 5\sqrt{3}$. Checking with Pythagoras: $5^2 + (5\sqrt{3})^2 = 25 + 75 = 100 = 10^2$.

Application: the shortest ramp

Leaning against a wall is a block of width $a$ and height $b$. A straight ramp must leave the floor, pass exactly through the top corner of the block and end at the wall. The shorter the ramp, the less covering material (marble, granite) it uses. Which slope $\alpha$ gives the shortest ramp?

The corner divides the ramp into two pieces, each the hypotenuse of a right-angled triangle. The top one has a horizontal leg $a$, adjacent to $\alpha$; the bottom one has a vertical leg $b$, opposite to $\alpha$. Then

$$L(\alpha) = \frac{a}{\cos\alpha} + \frac{b}{\operatorname{sen}\alpha}.$$

Ramping too much when standing or lying down becomes long; the best is in the middle. With derivatives, equating $L'(\alpha)$ to zero, we arrive at $\dfrac{a\operatorname{sen}\alpha}{\cos^2\alpha} = \dfrac{b\cos\alpha}{\operatorname{sen}^2\alpha}$, that is,

$$\operatorname{tg}^3\alpha^\star = \frac{b}{a} \qquad\Longrightarrow\qquad \operatorname{tg}\alpha^\star = \sqrt[3]{\frac{b}{a}}.$$
Change the slope and see the length $L(\alpha)$ on the graph. The green dot is the minimum, at $\operatorname{tg}\alpha^\star = \sqrt[3]{b/a}$.

ExemploA block 8 m wide and 1 m high

With $a = 8$ and $b = 1$: $\operatorname{tg}\alpha^\star = \sqrt[3]{\frac{1}{8}} = \frac{1}{2}$. Then $\cos\alpha^\star = \frac{2}{\sqrt{5}}$, $\operatorname{sen}\alpha^\star = \frac{1}{\sqrt{5}}$ and

$$L_{\min} = \frac{8\sqrt{5}}{2} + \frac{1 \cdot \sqrt{5}}{1} = 5\sqrt{5} \approx 11{,}18 \text{ m}.$$

The same calculation answers the classic question: what is the shortest ladder that, resting on the ground, passes over a wall of height $b$ and reaches a wall at a distance $a$ behind the wall?

The exercises in this section are numbers 352 to 359 from chapter list.