Exercises from chapter IX — Right triangles
Always draw a figure: mark the right angle, the data and the unknown. Then choose the right triangle in which the unknown and the data appear together.
Metric relationships
Calculate the elements $a$, $h$, $m$ and $n$ indicated in the figure (categories $b = 4$ and $c = 3$).
See the resolution
Being $c = 3$ and $b = 4$:
$$a^2 = b^2 + c^2 = 16 + 9 = 25 \Rightarrow a = 5$$$$h = \frac{bc}{a} = \frac{4 \cdot 3}{5} = \frac{12}{5}$$$$m = \frac{c^2}{a} = \frac{9}{5}$$$$n = \frac{b^2}{a} = \frac{16}{5}.$$Calculate the elements $x$, $y$, $z$, $t$ in the figure: the legs measure 12 and 5, $t$ is the hypotenuse, $x$ is the height relative to it and $y$ and $z$ are the projections of the legs 5 and 12 on the hypotenuse.
Ver resposta
$x = \dfrac{60}{13}$, $y = \dfrac{25}{13}$, $z = \dfrac{144}{13}$, $t = 13$
Write 6 metric relations involving the elements of the figure: a right-angled triangle with legs $\alpha$ and $\beta$ and hypotenuse $\gamma$, with height $\varphi$ relative to the hypotenuse, which is divided into projections $\delta$ (from $\alpha$) and $\varepsilon$ (from $\beta$).
Ver resposta
For example: $\alpha^2 = \gamma\delta$, $\beta^2 = \gamma\varepsilon$, $\varphi^2 = \delta\varepsilon$, $\alpha^2 + \beta^2 = \gamma^2$, $\alpha\beta = \gamma\varphi$, $\alpha^2 = \delta^2 + \varphi^2$
Calculate the area of a right-angled triangle in which the projections of the legs on the hypotenuse measure, respectively, 4 and 9.
Ver resposta
$39$
Calculate the hypotenuse of a right-angled triangle with perimeter 56 and height relative to the hypotenuse $\dfrac{168}{25}$.
Ver resposta
$25$
A isosceles triangle $ABC$ has a base $a = 12$ and is inscribed in a circle of diameter $2R = 20$. Calculate the measurements of the sides $b$ and $c$ of the triangle.
See the resolution
Connecting the vertex $A$ to the center $O$, we obtain a straight line that cuts the circle at $D$: $\overline{AD}$ is diameter. Since $AD$ is perpendicular bisector of $\overline{BC}$, we have $AD \perp BC$ and $BM = MC = 6$. The triangle $ADC$ is rectangular in $C$ (inscribed in a semicircle), and $\overline{CM}$ is its height. With $AM = y$ and $MD = 20 - y$:
$$6^2 = y(20 - y) \Longrightarrow y^2 - 20y + 36 = 0 \Longrightarrow y = 2 \;\text{ ou }\; y = 18.$$Then $b^2 = 6^2 + y^2$, that is, $b^2 = 40$ or $b^2 = 360$. There are two possible triangles: $b = c = 2\sqrt{10}$ (when $A$ and the center $O$ are on opposite sides of the base, $y = 2$) or $b = c = 6\sqrt{10}$ (when they are on the same side, $y = 18$).
Calculate the height of a isosceles triangle knowing the radius $R$ of the circumscribed circle and the base $a$. Data: $R = 5$ and $a = 8$.
Ver resposta
$h = 8$ or $h = 2$
Calculate the radius of the circle circumscribed by a isosceles triangle base 6 whose other sides measure $\sqrt{90}$.
Ver resposta
$5$
A point $P$ is distant $d = 13$ from the center $O$ of a circle of radius $R = 5$. If we draw through $P$ a tangent line to the circle at point $T$, what is the measurement of the segment $\overline{PT}$?
Ver resposta
$PT = 12$
Calculate the side of a regular octagon inscribed in a circle of radius $R$.
Ver resposta
$\ell_8 = R\sqrt{2 - \sqrt{2}}$
Trigonometric ratios
Calculate the internal angles of a right-angled triangle whose legs are $b = 9$ and $c = 12$.
See the resolution
Answer: $\hat{A} = 90^\circ$, $\hat{B} = \operatorname{arc\,tg}\frac{3}{4}$ (about $36^\circ 52'$) and $\hat{C} = \operatorname{arc\,tg}\frac{4}{3}$ (about $53^\circ 8'$).
Calculate the sides of a right triangle knowing that the height relative to the hypotenuse is $h = 4$ and an acute angle is $\hat{B} = 30^\circ$.
Ver resposta
$a = \dfrac{16\sqrt{3}}{3}$, $b = \dfrac{8\sqrt{3}}{3}$, $c = 8$
Calculate the sides of a right-angled triangle knowing that the height relative to the hypotenuse measures 4 and forms an angle of $15^\circ$ with the side $b$.
Ver resposta
$a = 16$, $b = \dfrac{16}{\sqrt{6} + \sqrt{2}}$, $c = \dfrac{16}{\sqrt{6} - \sqrt{2}}$
Calculate the legs of a right-angled triangle $ABC$ knowing that it is inscribed in a circle of radius 3 and has acute angles such as $\hat{B} = 2\hat{C}$.
Ver resposta
$b = 3\sqrt{3}$ e $c = 3$
Calculate the acute angles of a right-angled triangle with hypotenuse 20, knowing that the median relative to one of the legs measures 15.
Ver resposta
$\operatorname{arc\,tg}\sqrt{\dfrac{5}{7}}$ e $\operatorname{arc\,tg}\sqrt{\dfrac{7}{5}}$
(FFCLUSP-66) In the figure, the angles $O\hat{A}_iB_i$ and $O\hat{B}_{i+1}A_i$ ($i = 0, 1, 2, 3, \ldots$) are right angles. How much is the sum of the segments $\overline{A_0B_0}$, $\overline{A_1B_1}$, $\overline{A_2B_2}$, … worth in terms of $\overline{A_0B_0}$ and $\theta$?
View a script
In triangle $A_0B_1B_0$, rectangle in $B_1$, angle in $B_0$ measures $90^\circ - \theta$; logo $A_0B_1 = A_0B_0\cos\theta$. Likewise, in triangle $A_0A_1B_1$, rectangle in $A_1$, $A_1B_1 = A_0B_1\cos\theta = A_0B_0\cos^2\theta$. Each vertical segment is the previous one multiplied by $\cos^2\theta$: the sum is an infinite geometric progression of ratio $\cos^2\theta < 1$,
$$A_0B_0\,(1 + \cos^2\theta + \cos^4\theta + \cdots) = \frac{A_0B_0}{1 - \cos^2\theta} = \frac{A_0B_0}{\operatorname{sen}^2\theta} = A_0B_0\operatorname{cossec}^2\theta.$$Ver resposta
$A_0B_0 \cdot \operatorname{cossec}^2\theta$
Calculate the angle formed by the diagonal and the smallest side of a rectangle whose sides are in the ratio $\dfrac{3}{4}$.
Ver resposta
$\operatorname{arc\,tg}\dfrac{4}{3}$
Calculate the area of a isosceles triangle $ABC$ whose base is $a = 8$, knowing that $\hat{A} = 45^\circ$.
See the resolution
Plotting the height $\overline{AM}$, the triangle is divided into two congruent parts, with $\hat{A}' = \hat{A}'' = 22^\circ 30'$ and $BM = MC = 4$. Then
$$\operatorname{tg} 22^\circ 30' = \frac{MC}{AM} \Longrightarrow AM = \frac{4}{\operatorname{tg} 22^\circ 30'} = \frac{4\sqrt{2 + \sqrt{2}}}{\sqrt{2 - \sqrt{2}}},$$using $\operatorname{tg} 22^\circ 30' = \sqrt{\frac{1 - \cos 45^\circ}{1 + \cos 45^\circ}} = \frac{\sqrt{2 - \sqrt{2}}}{\sqrt{2 + \sqrt{2}}}$ (arc half). The area is
$$S = \frac{BC \cdot AM}{2} = \frac{16\sqrt{2 + \sqrt{2}}}{\sqrt{2 - \sqrt{2}}} = 16(\sqrt{2} + 1),$$because $\frac{\sqrt{2 + \sqrt{2}}}{\sqrt{2 - \sqrt{2}}} = \frac{2 + \sqrt{2}}{\sqrt{4 - 2}} = \sqrt{2} + 1$.
Calculate the height relative to the base of a isosceles triangle with perimeter $2p = 36$, knowing that the angles adjacent to the base are equal to $\operatorname{arc\,cos}\dfrac{2}{7}$.
See the resolution
Let $a$ be the base and $b$ each of the sides equal. The height divides the base in half, and $\frac{a}{2} = b\cos\hat{B} = \frac{2b}{7}$, that is, $a = \frac{4b}{7}$. From the perimeter, $\frac{4b}{7} + 2b = 36$, then $b = 14$ and $a = 8$. Then $h = \sqrt{14^2 - 4^2} = \sqrt{180} = 6\sqrt{5}$.
Ver resposta
$h = 6\sqrt{5}$ (sides measure 14, 14 and 8). The printed template shows $2\sqrt{37}$, a value that is not compatible with the data.
Calculate the radius of the circle circumscribed to a isosceles triangle base $a = 8$ and angle opposite the base $\hat{A} = 120^\circ$.
Ver resposta
$R = \dfrac{8\sqrt{3}}{3}$
A straight line determines, on a circle of radius 10, a chord of length 16. What is the measurement of the central angle from which the rope is seen?
Ver resposta
$2\operatorname{arc\,sen}\dfrac{4}{5}$
Determine the angle $\hat{B}$ of a right triangle $ABC$ in $A$ knowing that the relationship $\dfrac{1}{c} + \dfrac{2}{b} = \dfrac{\sqrt{5}}{h}$ is verified, where $h$ is the height relative to the hypotenuse.
Ver resposta
$\hat{B} = \operatorname{arc\,tg}\dfrac{1}{2} \approx 26^\circ 34'$
In a triangle $ABC$ right angled to $A$, it is known that the acute angle formed by the medians $\overline{BM}$ and $\overline{CN}$ is $\theta = 30^\circ$. Calculate the angle $\hat{C}$.
Ver resposta
$\hat{C} = \dfrac{1}{2}\operatorname{arc\,sen}\dfrac{4}{3\sqrt{3}}$ (or the complement of that angle, as the roles of $B$ and $C$ can be interchanged)
In a right triangle $ABC$ in $A$ the internal angle bisectors $\overline{BB'}$ and $\overline{CC'}$ are drawn. Knowing that $AB' = 1$ and $AC' = 1$, calculate the angle $\hat{B}$ and the hypotenuse $a$.
Ver resposta
$\hat{B} = 2\operatorname{arc\,tg}(\sqrt{2} - 1) = 45^\circ$ and $a = 2 + \sqrt{2}$. The printed template gives $a = 2\sqrt{2}$; but, with $\hat{B} = 45^\circ$, $c = \frac{1}{\operatorname{tg} 22^\circ 30'} = \sqrt{2} + 1$ and $a = c\sqrt{2} = 2 + \sqrt{2}$.
Calculate the acute angles of the right triangle with hypotenuse $a = 13$ m, knowing that the radius of the inscribed circle is $r = 2$ m.
Ver resposta
$\operatorname{arc\,sen}\dfrac{5}{13}$ e $\operatorname{arc\,sen}\dfrac{12}{13}$
An observer sees a building, built on level ground, at an angle of $60^\circ$. Moving further away from the building 30 m, you can see it from an angle of $45^\circ$. How tall is the building?
See the resolution
In triangle $BXY$: $\operatorname{tg} 60^\circ = \dfrac{h}{\ell} \Rightarrow \ell = \dfrac{h}{\sqrt{3}}$. In triangle $AXY$: $\operatorname{tg} 45^\circ = \dfrac{h}{\ell + 30} \Rightarrow h = \ell + 30$. Then
$$h = \frac{h}{\sqrt{3}} + 30 \Longrightarrow h = \frac{30\sqrt{3}}{\sqrt{3} - 1} = 15\sqrt{3}(\sqrt{3} + 1) = 45 + 15\sqrt{3} \approx 71 \text{ m}.$$(MAUÁ-68) To measure the height of the vertical tower $\overline{DE}$, the segment $\overline{AB}$ of length 12 m and whose midpoint is $C$ is taken in the horizontal plane that passes through its base $D$. The angles $D\hat{A}E$, $D\hat{B}E$ and $D\hat{C}E$ are measured, verifying that $D\hat{A}E = D\hat{B}E = 45^\circ$ and $D\hat{C}E = 60^\circ$. Determine the height of the tower.
Ver resposta
$h = 3\sqrt{6}$ m
Calculate the distance between the sills of two windows of a skyscraper, knowing the angles $\alpha$ and $\beta$ ($\alpha < \beta$) under which they are observed from a point $O$ on the ground and the distance $d$ from that point to the building.
Ver resposta
$h = d(\operatorname{tg}\beta - \operatorname{tg}\alpha)$
(MAUÁ-65) To obtain the height $H$ of a chimney $\overline{CD}$, an engineer, with a special device, established the horizontal $\overline{AB}$ ($B$ in the chimney) and measured the angles $\alpha = B\hat{A}C$ (down, to the base $C$) and $\beta = B\hat{A}D$ (up, to the top $D$), having the follow measurement $BC = h$. Determine the height of the chimney.
Ver resposta
$H = h\left(\dfrac{\operatorname{tg}\beta}{\operatorname{tg}\alpha} + 1\right)$
(MAUÁ-67) An observer is on a highway, in a straight and horizontal section located on the same vertical plane that contains a TV tower at the top of a peak. From two positions $A$ and $B$ of this section, distant 60 m from each other, the observer sees the upper end of the tower, respectively, under the angles of $30^\circ$ and $31^\circ 53'$. The device that measures the angles was placed 1,50 m above the track, which is 721,50 m above sea level. Determine the height of the end of the tower in relation to sea level. Data: $\operatorname{tg} 31^\circ 53' \approx 0{,}62$.
View a script
Let $x$ be the horizontal distance from $B$ to the tower and $t$ be the height of the top above the device. Then $t = 0{,}62\,x$ and $t = (x + 60)\operatorname{tg} 30^\circ$. With $\operatorname{tg} 30^\circ = \frac{\sqrt{3}}{3} \approx 0{,}5774$:
$$0{,}62x = 0{,}5774x + 34{,}64 \Longrightarrow x \approx 812 \text{ m} \Longrightarrow t \approx 504 \text{ m}.$$The height above sea level is $721{,}50 + 1{,}50 + 504 \approx 1227$ m.
Ver resposta
About $1227$ m (the printed template gives $1233$ m; the result is very sensitive to rounding of $\operatorname{tg} 30^\circ$, because the difference $0{,}62 - \operatorname{tg} 30^\circ$ is small)
(LINS-66) In the figure, $A$, $C$ and $B$ are aligned on the floor with $AC = CB = 60$ m; $\overline{AM} = x$ and $\overline{BN} = y$ are vertical; $N\hat{A}B = 2\alpha$, $M\hat{B}A = \alpha$ and $M\hat{C}N = 90^\circ$. Calculate the distances $x$ and $y$.
View a script
The triangles $MAC$ and $CBN$ are similar (the angles $M\hat{C}A$ and $C\hat{N}B$ are equal, as they both complete $N\hat{C}B$ to $90^\circ$): $\frac{x}{60} = \frac{60}{y}$, that is, $xy = 3600$. Also, $\operatorname{tg}\alpha = \frac{x}{120}$ and $\operatorname{tg} 2\alpha = \frac{y}{120}$. By the double arc formula,
$$\frac{y}{120} = \frac{2 \cdot \frac{x}{120}}{1 - \frac{x^2}{14\,400}} \Longrightarrow \frac{3600}{x} = \frac{28\,800x}{14\,400 - x^2} \Longrightarrow 14\,400 - x^2 = 8x^2 \Longrightarrow x = 40,$$e $y = \frac{3600}{40} = 90$.
Ver resposta
$x = 40$ m e $y = 90$ m
Solving right-angled triangles
Solve a right triangle $ABC$ knowing the measure of the internal angle bisector $S_b = 5$ and the angle $\hat{C} = 30^\circ$.
See the resolution
It is immediate that $\hat{B} = 60^\circ$ and $\frac{\hat{B}}{2} = 30^\circ$. In the right triangle $ABS$ ($S$ is the foot of angle bisector, over $\overline{AC}$):
$$c = 5\cos 30^\circ = \frac{5\sqrt{3}}{2}.$$Then $a = \dfrac{c}{\cos\hat{B}} = \dfrac{5\sqrt{3}/2}{1/2} = 5\sqrt{3}$ and $b = a\operatorname{sen}\hat{B} = 5\sqrt{3} \cdot \dfrac{\sqrt{3}}{2} = \dfrac{15}{2}$.
Solve a right triangle $ABC$ given $b = 3$ and $a - c = \sqrt{3}$.
Ver resposta
$\hat{B} = 60^\circ$, $\hat{C} = 30^\circ$, $a = 2\sqrt{3}$ e $c = \sqrt{3}$
Solve a right triangle $ABC$ in $A$ knowing that $a + b = 18$ and $a + c = 25$.
Ver resposta
$a = 13$, $b = 5$, $c = 12$, $\hat{B} = \operatorname{arc\,sen}\dfrac{5}{13}$, $\hat{C} = \operatorname{arc\,sen}\dfrac{12}{13}$
Solve a right triangle $ABC$ knowing that $a = 4$ and that the measure of angle bisector inside $\overline{BE}$ is $S_b = 6\sqrt{2} - 2\sqrt{6}$.
Ver resposta
$\hat{B} = 30^\circ$, $\hat{C} = 60^\circ$, $b = 2$, $c = 2\sqrt{3}$
Solve a right triangle knowing the height $h = 1$ relative to the hypotenuse and the perimeter $2p = 2\sqrt{2} + 2$.
Ver resposta
$a = 2$, $b = c = \sqrt{2}$, $\hat{B} = \hat{C} = \dfrac{\pi}{4}$
Solve a isosceles triangle $ABC$ knowing that the height relative to the base $\overline{BC}$ measures $h = 24$ and the perimeter is $2p = 64$.
Ver resposta
$a = 14$, $b = c = 25$, $\hat{A} = 2\operatorname{arc\,sen}\dfrac{7}{25}$, $\hat{B} = \hat{C} = \operatorname{arc\,cos}\dfrac{7}{25}$
Solve a right triangle $ABC$ knowing the radius $r = 2$ of the inscribed circle and the height $h = \dfrac{60}{13}$ relative to the hypotenuse.
Ver resposta
$a = 13$, $b = 12$, $c = 5$, $\hat{B} = \operatorname{arc\,sen}\dfrac{12}{13}$, $\hat{C} = \operatorname{arc\,sen}\dfrac{5}{13}$
Solve a right-angled triangle $ABC$ knowing the height $h = 2$ relative to the hypotenuse and the radius $r' = 2\sqrt{2} + 2$ of the ex-inscribed circle located at the right angle.
Ver resposta
$a = 4$, $b = c = 2\sqrt{2}$, $\hat{B} = \hat{C} = \dfrac{\pi}{4}$