Right triangles and any triangles: trigonometric ratios, laws of sines and cosines (chapters IX and X).
Teste 189CESCEA-74
Among the right-angled triangles below, one and only one presents the correct data. Mark it:
In (c), $\cos 60^\circ = \frac{3}{6} = \frac{1}{2}$: the adjacent side is half the hypotenuse. In (a) the side adjacent to $30^\circ$ should be $4\cos 30^\circ = 2\sqrt{3}$; in (b) the angle opposite the side $\sqrt{3}$ has tangent $\frac{\sqrt{3}}{3}$ and measures $30^\circ$; in (d) the sides of a triangle with an angle of $45^\circ$ would be equal.
Teste 190CESCEM-75
Considering the triangle $ABC$, a rectangle in $A$, with $a = 7{,}5$ m, $b = 4{,}5$ m and $c = 6$ m, with $x$ being the angle at the vertex $B$, it can be stated that $\operatorname{tg} x$ is equal to:
$x$ is the angle at $B$; the opposite side is $b = 4{,}5$ and the adjacent one, $c = 6$: $\operatorname{tg} x = \frac{4{,}5}{6} = 0{,}75$.
Teste 191CESCEA-77
In the triangle $ABC$, a rectangle in $A$, $BC = 10$ and $\cos\alpha = \frac{3}{5}$ are given, with $\alpha$ being the angle in $B$. The sum of the legs is:
$AB = BC\cos\alpha = 10 \cdot \frac{3}{5} = 6$ and $AC = BC\operatorname{sen}\alpha = 10 \cdot \frac{4}{5} = 8$; the sum is 14.
Teste 192MACK-77
In the figure, the rectangle has height 50; the straight line that goes from the bottom left corner to $A$ forms $60^\circ$ with the base, and the one that goes from the top left corner to $A$ forms $30^\circ$ with the top side. $A$ is vertical to $B$. $AB$ is worth:
Let $w$ be the width of the rectangle and $h = AB$. By the angle of $60^\circ$ in the lower left corner, $h = w\sqrt{3}$; by the angle of $30^\circ$ in the upper left corner, $h - 50 = \frac{w}{\sqrt{3}}$. Subtracting, $w\left(\sqrt{3} - \frac{1}{\sqrt{3}}\right) = 50$, $w = 25\sqrt{3}$ and $h = 75$.
Teste 193EPUSP-66
$\overline{AB}$ is the hypotenuse of a right triangle $ABC$. The median $\overline{AD}$ measures 7 and the median $\overline{BE}$ measures 4. The length of $\overline{AB}$ is equal to:
With the right angle at $C$, sides $a = BC$ and $b = AC$: $AD^2 = b^2 + \frac{a^2}{4} = 49$ and $BE^2 = a^2 + \frac{b^2}{4} = 16$. Adding, $\frac{5}{4}(a^2 + b^2) = 65$, then $AB^2 = a^2 + b^2 = 52$ and $AB = 2\sqrt{13}$.
Teste 194GV-70
In the triangle $ABC$ we know that $\hat{A} = 90^\circ$, $\hat{B} = 60^\circ$ and $AB = 50$ cm. Then segment $\overline{AC}$ measures:
$AC$ is the opposite side of $\hat{B}$: $AC = AB\operatorname{tg} 60^\circ = 50\sqrt{3}$ cm.
Teste 195CESCEM-76
A person 1,70 m tall observes the top of a tree at an angle $\alpha$ with the horizontal; $b$ is the horizontal distance from the eyes to the trunk and $a$ is the distance from the eyes to the top. To know approximately the height of the tree, you must add 1,70 m with:
Above eye level, the height is the leg opposite $\alpha$ in the adjacent leg triangle $b$ (the horizontal distance): $b\operatorname{tg}\alpha$. Add the height of the eyes, 1,70 m.
Teste 196CESCEA-76
In the figure, $ABC$ is rectangle in $B$, $D$ is between $A$ and $B$, $AD = d$, $BC = h$, $C\hat{A}D = \alpha$ and $C\hat{D}B = \beta$. Then:
In right triangles $ABC$ and $DBC$: $AB = h\operatorname{cotg}\alpha$ and $DB = h\operatorname{cotg}\beta$. Then $d = AB - DB = h(\operatorname{cotg}\alpha - \operatorname{cotg}\beta)$.
Teste 197CESCEA-73
In the triangle $ABC$ there are $b = 2$, $\hat{B} = 45^\circ$ and $\hat{C} = 60^\circ$. Then the $a$ side measures:
$\hat{A} = 75^\circ$ and, according to the law of sines, $a = \frac{b\operatorname{sen}\hat{A}}{\operatorname{sen}\hat{B}} = \frac{2 \cdot \frac{\sqrt{6} + \sqrt{2}}{4}}{\frac{\sqrt{2}}{2}} = \sqrt{3} + 1$.
Teste 198CESCEM-72
In a right-angled triangle in which one side is worth 1 and the other is worth $\operatorname{tg}\varphi$, the hypotenuse is worth:
The hypotenuse is $\sqrt{1 + \operatorname{tg}^2\varphi} = \sqrt{\sec^2\varphi} = |\sec\varphi|$ (a length is always positive).
Teste 199GV-73
Consider the triangle $ABC$, a rectangle in $A$, with hypotenuse $a$ and legs $b = AC$ and $c = AB$, and indicate its area by $S$. Select the true statement:
In the figure, $c$ is a circle of radius $R$, $r$ and $s$ are tangent lines to the circle that intersect at $O$, $T$ is the point of tangency at $s$ and $OT = 2R$. Then the angle $\alpha$ of the straight lines $r$ and $s$ checks:
The center of the circle is at angle bisector of angle $\alpha$: $\operatorname{tg}\frac{\alpha}{2} = \frac{R}{2R} = \frac{1}{2}$. Then $\operatorname{tg}\alpha = \frac{2 \cdot \frac{1}{2}}{1 - \frac{1}{4}} = \frac{4}{3}$, that is, $\operatorname{sen}\alpha = \frac{4}{5}$ and $\cos\alpha = \frac{3}{5}$.
Teste 201SANTA CASA-73
In a circle with center $O$ and radius $R$, a measuring chord $\frac{3R}{2}$ determines the central angle $\alpha$. Regarding $\alpha$, it can be said that:
The measuring chord $\frac{3R}{2}$ subtends the central angle $\alpha$: $\operatorname{sen}\frac{\alpha}{2} = \frac{3R/4}{R} = \frac{3}{4}$, $\cos\frac{\alpha}{2} = \frac{\sqrt{7}}{4}$ and $\operatorname{sen}\alpha = 2 \cdot \frac{3}{4} \cdot \frac{\sqrt{7}}{4} = \frac{3\sqrt{7}}{8}$. (In the book the alternatives are labeled a, b, c, e — without d — and the template indicates d; the correct value is that of the third alternative.)
Teste 202MAUÁ-68
In a triangle $ABC$ it is assumed that $2\operatorname{tg}\hat{A} = \operatorname{tg}\hat{B} + \operatorname{tg}\hat{C}$ and $0 < \hat{A} < \frac{\pi}{2}$. In this triangle the relationship applies:
In a triangle (without a right angle), $\operatorname{tg}\hat{A} + \operatorname{tg}\hat{B} + \operatorname{tg}\hat{C} = \operatorname{tg}\hat{A}\operatorname{tg}\hat{B}\operatorname{tg}\hat{C}$, because $\operatorname{tg}(\hat{B} + \hat{C}) = -\operatorname{tg}\hat{A}$. With $\operatorname{tg}\hat{B} + \operatorname{tg}\hat{C} = 2\operatorname{tg}\hat{A}$: $3\operatorname{tg}\hat{A} = \operatorname{tg}\hat{A}\operatorname{tg}\hat{B}\operatorname{tg}\hat{C}$, and as $\operatorname{tg}\hat{A} \ne 0$, $\operatorname{tg}\hat{B}\operatorname{tg}\hat{C} = 3$.
Teste 203ITA-77
Consider a triangle $ABC$ whose internal angles $\hat{A}$, $\hat{B}$ and $\hat{C}$ verify the relationship $\operatorname{sen}\hat{A} = \operatorname{tg}\frac{\hat{B} + \hat{C}}{2}$. So we can say that:
In a triangle $ABC$, the angles $\hat{A}$ and $\hat{B}$ measure, respectively, $60^\circ$ and $45^\circ$; the side $\overline{BC}$ measures $5\sqrt{6}$ cm. So the measurement of side $\overline{AC}$ is:
By the law of sines, $AC = \frac{BC\operatorname{sen}\hat{B}}{\operatorname{sen}\hat{A}} = \frac{5\sqrt{6} \cdot \frac{\sqrt{2}}{2}}{\frac{\sqrt{3}}{2}} = 5\sqrt{\frac{12}{3}} = 10$ cm.
Teste 205ITA-73
A ship, sailing in a straight line, passes successively through the points $A$, $B$, $C$. When the ship is at $A$, the captain observes a beacon $L$ and calculates the angle $L\hat{A}C = 30^\circ$. After navigating 4 miles to $B$, check the $L\hat{B}C = 75^\circ$ angle. How many miles separate the lighthouse from point $B$?
In the triangle $ABL$: the angle at $A$ measures $30^\circ$, the angle at $B$ measures $180^\circ - 75^\circ = 105^\circ$, and therefore the angle at $L$ measures $45^\circ$. By the law of sines, $LB = \frac{AB\operatorname{sen} 30^\circ}{\operatorname{sen} 45^\circ} = \frac{4 \cdot \frac{1}{2}}{\frac{\sqrt{2}}{2}} = 2\sqrt{2}$.
Teste 206MACK-76
In the figure, $\overline{AC} \perp \overline{CB}$, $\overline{AD} \perp \overline{DC}$, $m(D\hat{A}C) = m(A\hat{B}C) = \beta$ and $AB = a$. The value of $AD$, as a function of $a$ and $\beta$, is:
In triangle $ACB$, rectangle in $C$: $AC = a\operatorname{sen}\beta$. In triangle $ADC$, rectangle in $D$, with angle $\beta$ in $A$: $AD = AC\cos\beta = a\operatorname{sen}\beta\cos\beta = \frac{1}{2}a\operatorname{sen} 2\beta$.
Teste 207ITA-74
It is desired to build a railway connecting point $A$ to point $B$, which is $40\sqrt{2}$ km southeast of $A$. A lake, on the plain where $A$ and $B$ are, prevents construction in a straight line. To bypass the lake, the road will be built in 2 straight sections with apex at point $C$, which is 36 km east and 27 km south of $A$. The length of the section $\overline{CB}$ is:
With $A$ at the origin, axis $x$ to the east and $y$ to the north: $B = (40, -40)$ (at $40\sqrt{2}$ km in the southeast direction) and $C = (36, -27)$. Then $CB = \sqrt{4^2 + 13^2} = \sqrt{185}$ km.
Teste 208EPUSP-66
The sides of a triangle are in the ratio $6 : 8 : 9$. Then:
The square of the longest side, $9^2 = 81$, is smaller than $6^2 + 8^2 = 100$: the triangle is acute (exercise 372).
Teste 209FFCLUSP-67
Given the segments $\overline{AC}$ and $\overline{BC}$ and an angle $\hat{B}$, it is possible to construct a triangle that has $\overline{AC}$ and $\overline{BC}$ as sides and $\hat{B}$ as an angle not adjacent to $\overline{AC}$ when:
It is the ambiguous case, with the side $AC$ opposite the angle $\hat{B}$ and the side $BC$ adjacent to it. If $AC > BC$, there is always a (single) triangle, whatever $\hat{B}$ is (with $0 < \hat{B} < \pi$). In other conditions, there may be no triangle.
Teste 210ITA-75
In a scalene triangle $ABC$, the sides opposite the angles $\hat{A}$, $\hat{B}$, $\hat{C}$ measure, respectively, $a$, $b$, $c$. Then the expression $a\operatorname{sen}(\hat{B} - \hat{C}) + b\operatorname{sen}(\hat{C} - \hat{A}) + c\operatorname{sen}(\hat{A} - \hat{B})$ has a value that satisfies one of the following alternatives:
It is relation I of exercise 399: with $a = 2R\operatorname{sen}(\hat{B} + \hat{C})$ etc., each term becomes a difference of squares of sines, and the sum is zero.
Teste 211GV-75
The side of the regular octagon inscribed in a circle of unit radius is $\sqrt{2 - \sqrt{2}}$. It can be concluded that $\cos\frac{\pi}{8}$ is worth:
The side of the octagon is the chord of $45^\circ$: $2\operatorname{sen}\frac{\pi}{8} = \sqrt{2 - \sqrt{2}}$. Then $\operatorname{sen}^2\frac{\pi}{8} = \frac{2 - \sqrt{2}}{4}$, $\cos^2\frac{\pi}{8} = \frac{2 + \sqrt{2}}{4}$ and $\cos\frac{\pi}{8} = \frac{\sqrt{2 + \sqrt{2}}}{2}$.
Teste 212MACK-74
The base $\overline{AD}$ of a rectangle, which is three times greater than its height $\overline{AB}$, is subdivided by the points $M$ and $N$ into three parts of equal size. Under these conditions, $A\hat{M}B + A\hat{N}B + A\hat{D}B$ is equal to:
With height $AB = h$: $AM = h$, $AN = 2h$, $AD = 3h$. Then $A\hat{M}B = 45^\circ$, $\operatorname{tg} A\hat{N}B = \frac{1}{2}$ and $\operatorname{tg} A\hat{D}B = \frac{1}{3}$. As $\operatorname{arc\,tg}\frac{1}{2} + \operatorname{arc\,tg}\frac{1}{3} = 45^\circ$ (exercise 268), the sum is $90^\circ$.
Teste 213ITA-75
Let $ABCD$ be a convex quadrilateral inscribed in a circle. It is known that $\hat{A} = 2\hat{C}$, $\hat{B} > \hat{D}$ and $\operatorname{tg}\hat{B}\operatorname{tg}\hat{D} + \operatorname{sen}\hat{A}\operatorname{sen}\hat{C} = -\frac{9}{4}$. In this case, the values of $\hat{A}$, $\hat{B}$, $\hat{C}$, $\hat{D}$ are, respectively:
In the inscribed quadrilateral, $\hat{A} + \hat{C} = 180^\circ$ and $\hat{B} + \hat{D} = 180^\circ$. With $\hat{A} = 2\hat{C}$: $\hat{C} = 60^\circ$, $\hat{A} = 120^\circ$ and $\operatorname{sen}\hat{A}\operatorname{sen}\hat{C} = \frac{3}{4}$. Like $\operatorname{tg}\hat{D} = -\operatorname{tg}\hat{B}$: $-\operatorname{tg}^2\hat{B} + \frac{3}{4} = -\frac{9}{4}$, $\operatorname{tg}^2\hat{B} = 3$; with $\hat{B} > \hat{D}$, $\hat{B} = 120^\circ$ and $\hat{D} = 60^\circ$.
Teste 214ITA-77
Let $A$, $B$ and $C$ be three distinct points on a straight line, with $B$ between $A$ and $C$. Let $a$ and $b$ ($a > 2b$) be the lengths of $\overline{AB}$ and $\overline{BC}$, respectively. If the segment $\overline{BD}$ is perpendicular to the segment $\overline{AC}$, how much must $BD = x$ measure so that the angle $B\hat{D}C$ is half of $B\hat{D}A$?
With $BD = x$ and $B\hat{D}C = \theta$: $\operatorname{tg}\theta = \frac{b}{x}$ and $\operatorname{tg} 2\theta = \frac{a}{x}$. By the double arc, $\frac{a}{x} = \frac{2b/x}{1 - b^2/x^2}$, that is, $a - \frac{ab^2}{x^2} = 2b$, $x^2 = \frac{ab^2}{a - 2b}$ and $x = \frac{ab}{\sqrt{a(a - 2b)}}$.
Teste 215ITA-77
Let $d$ and $L$, respectively, be the lengths of the diagonal $\overline{BD}$ and the side $\overline{BC}$ of the parallelogram $ABCD$. The diagonal $\overline{BD}$ forms the angle $\alpha$ with $\overline{BC}$ and the angle $\beta$ with $\overline{BA}$. The length $x$ of side $\overline{AB}$ is given by:
In the triangle $ABD$: $AB = x$, $BD = d$, $A\hat{B}D = \beta$ and $A\hat{D}B = D\hat{B}C = \alpha$ (internal alternations, as $AD \parallel BC$); logo $B\hat{A}D = \pi - (\alpha + \beta)$. By the law of sines, $\frac{x}{\operatorname{sen}\alpha} = \frac{d}{\operatorname{sen}(\alpha + \beta)}$.