Exercises from chapter X — Any triangles

Calling $b = 8$ and $c = 12$ the given sides and $\hat{A} = 120^\circ$ the angle between them, the law of cosines gives

then $a = \sqrt{304} = 4\sqrt{19}$ m.

From the law of cosines, $\cos\hat{A} = \dfrac{b^2 + c^2 - a^2}{2bc} = \dfrac{169 + 225 - 100}{2 \cdot 13 \cdot 15} = \dfrac{294}{390} = \dfrac{49}{65}$; therefore $\hat{A} = \operatorname{arc\,cos}\dfrac{49}{65}$.

By the law of cosines, $\cos\hat{A} = \dfrac{b^2 + c^2 - a^2}{2bc}$. If $a$, $b$, $c$ are rational, the numerator and denominator are rational (sums and products of rationals), and the denominator is not null; therefore $\cos\hat{A}$ is rational. The same goes for $\cos\hat{B}$ and $\cos\hat{C}$. (The sines, however, can be irrational: in equilateral triangle aside 1, $\operatorname{sen} 60^\circ = \frac{\sqrt{3}}{2}$.)

(So ​​that $b$ and $c$ are positive, $x > 1$.) Let us calculate $b^2 + c^2 + bc$:

Comparing with the law of cosines, $a^2 = b^2 + c^2 - 2bc\cos\hat{A}$, comes $-2bc\cos\hat{A} = bc$, that is, $\cos\hat{A} = -\frac{1}{2}$ and $\hat{A} = 120^\circ$.

As $2\operatorname{sen}^2\frac{\hat{C}}{2} = 1 - \cos\hat{C}$ and, in the right triangle, $\cos\hat{C} = \frac{b}{a}$:

Let us assume that $a$ is the largest side, that is, $a \ge b$ and $a \ge c$. The largest side is opposed to the largest angle, therefore $\hat{A} \ge \hat{B}$ and $\hat{A} \ge \hat{C}$, and the triangle is right-angled, acute-angled or obtuse-angled depending on whether $\hat{A}$ is straight, acute or obtuse. From the law of cosines, $\cos\hat{A} = \dfrac{b^2 + c^2 - a^2}{2bc}$. Then:

• a) $\hat{A} = 90^\circ \iff \cos\hat{A} = 0 \iff a^2 = b^2 + c^2$;
• b) $0^\circ < \hat{A} < 90^\circ \iff \cos\hat{A} > 0 \iff a^2 < b^2 + c^2$;
• c) $90^\circ < \hat{A} < 180^\circ \iff \cos\hat{A} < 0 \iff a^2 > b^2 + c^2$.

Conclusion: a triangle is right-angled, acute-angled or obtuse-angled depending on whether the square of its largest side is equal to, less than or greater than the sum of the squares of the other two.

From the law of sines, $2R = \dfrac{a}{\operatorname{sen}\hat{A}} = \dfrac{15}{\frac{1}{2}} = 30$ cm; then $R = 15$ cm.

$\hat{A} = 180^\circ - 30^\circ - 45^\circ = 105^\circ$, and $\operatorname{sen} 105^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}$. By the law of sines:

Rationalizing: $b = 5(\sqrt{6} - \sqrt{2})$ and $c = 10(\sqrt{3} - 1)$.

a) Construct any triangle $A'B'C'$ with angles $\alpha$ and $\beta$ (it has the right shape, but not the size). On a ray, mark the perimeter of $A'B'C'$ and the perimeter $2p$ given; by Thales’ theorem (homothety), reduce or enlarge $A'B'C'$ in the ratio between the two perimeters.

b) By the law of sines, $a = 2R\operatorname{sen}\alpha$, $b = 2R\operatorname{sen}\beta$ and $c = 2R\operatorname{sen}(\alpha + \beta)$ (because $\hat{C} = 180^\circ - \alpha - \beta$). Adding, $2p = 2R[\operatorname{sen}\alpha + \operatorname{sen}\beta + \operatorname{sen}(\alpha + \beta)]$; hence $2R$ and the three sides.

As the angles $A\hat{D}B$ and $A\hat{C}B$ “see” the segment $\overline{AB}$ from the same angle, on the same side, the points $A$, $B$, $C$, $D$ are on the same circle, with radius $R$. By the law of sines in the triangle $ABD$, $AB = 2R\operatorname{sen} A\hat{D}B$; in the triangle $BCD$, $CD = 2R\operatorname{sen} C\hat{B}D$. Dividing member by member, $\dfrac{AB}{CD} = \dfrac{\operatorname{sen} A\hat{D}B}{\operatorname{sen} C\hat{B}D}$, which is the thesis.

The average base $\overline{C'B'}$ is parallel to $\overline{BC}$ and measures $\frac{a}{2}$; the distance between these two parallels is half the height $h_a$. As $C'M$ and $B'M$ make equal angles with $BC$, they make equal (alternate) angles with $C'B'$: the triangle $C'MB'$ is isosceles, and its height, $\frac{h_a}{2}$, falls in the middle of $\overline{C'B'}$. Then

By the law of sines, $a = 2R\operatorname{sen}\alpha$; and $h_a = c\operatorname{sen}\beta = 2R\operatorname{sen}\gamma\operatorname{sen}\beta$. Logo $\operatorname{cotg}\varphi = \dfrac{2R\operatorname{sen}\alpha}{4R\operatorname{sen}\beta\operatorname{sen}\gamma} = \dfrac{\operatorname{sen}\alpha}{2\operatorname{sen}\beta\operatorname{sen}\gamma}$.

In the triangle $ABY$, the external angle at $A$, which measures $\alpha$, is equal to the sum of the non-adjacent internal ones, $\frac{\alpha}{2}$ and $B\hat{Y}A$; then $B\hat{Y}A = \frac{\alpha}{2}$ and the triangle $ABY$ is isosceles: $AY = AB = 50$ m. In the right triangle $AXY$:

Let $\alpha$, $2\alpha$ and $3\alpha$ be the three angles, seen from the points $A$, $B$ and $C$, and $Y$ be the top of the tower. As in exercise 386, the triangle $ABY$ is isosceles and $BY = 110$ m. In the triangle $BCY$, the angle at $Y$ measures $3\alpha - 2\alpha = \alpha$ and the angle at $C$ measures $\pi - 3\alpha$; by the law of sines,

The height is $BY\operatorname{sen} 2\alpha = 110 \cdot 2 \cdot \frac{1}{\sqrt{5}} \cdot \frac{2}{\sqrt{5}} = 88$ m.

With $\cos 2x = 1 - 2\operatorname{sen}^2 x$:

According to the law of sines, $a\operatorname{sen}\hat{B} = b\operatorname{sen}\hat{A}$, and the parenthesis is null.

In the right triangle, $S_b = \frac{c}{\cos(\hat{B}/2)}$ and $S_c = \frac{b}{\cos(\hat{C}/2)}$, with $b = a\operatorname{sen}\hat{B}$ and $c = a\cos\hat{B}$. Then

For $\hat{B} = 60^\circ$: $\dfrac{\frac{\sqrt{3}}{4}}{\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{6} + \sqrt{2}}{4}} = \dfrac{2}{\sqrt{6} + \sqrt{2}} = \dfrac{\sqrt{6} - \sqrt{2}}{2}$. The expression is symmetric in $\hat{B}$ and $\hat{C}$ and decreases when $\hat{B}$ moves away from $45^\circ$, so the only solutions are $60^\circ$ and $30^\circ$.

a) The chord $AP$ subtends the central angle $2\alpha$: $AP = 2r\operatorname{sen}\alpha$. And $HP$ is the abscissa of $P$: $HP = r\cos 2\alpha = r(1 - 2\operatorname{sen}^2\alpha)$. The condition $AP - HP = r$ becomes $2\operatorname{sen}\alpha - 1 + 2\operatorname{sen}^2\alpha = 1$, that is, $\operatorname{sen}^2\alpha + \operatorname{sen}\alpha - 1 = 0$, or $1 - \operatorname{sen}^2\alpha = \operatorname{sen}\alpha$, or $\cos^2\alpha = \operatorname{sen}\alpha$, or even $\cos\alpha = \frac{\operatorname{sen}\alpha}{\cos\alpha} = \operatorname{tg}\alpha$.

b) From the quadratic equation, $\operatorname{sen}\alpha = \frac{-1 + \sqrt{5}}{2}$ (the other root is smaller than $-1$).

c) $\operatorname{sen}\alpha \approx 0{,}618034$, between $\operatorname{sen} 38^\circ 10'$ and $\operatorname{sen} 38^\circ 11'$. Interpolating: $\frac{0{,}618034 - 0{,}617951}{0{,}618180 - 0{,}617951} \cdot 60'' = \frac{83}{229} \cdot 60'' \approx 22''$. Logo $\alpha \approx 38^\circ 10' 22''$.

a) $S = \frac{ab}{2}\operatorname{sen}\theta$, to $0 < \theta < \pi$: a sinusoid arc of height $\frac{ab}{2}$ (in the graph, $\frac{ab}{2} = 1$).

b) The maximum occurs at $\theta = 90^\circ$.

c) $\frac{\operatorname{sen} 120^\circ}{\operatorname{sen} 30^\circ} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3} \approx 1{,}732$: an increase of approximately $73\%$.

In all of them we use $a = 2R\operatorname{sen}\hat{A}$, $b = 2R\operatorname{sen}\hat{B}$, $c = 2R\operatorname{sen}\hat{C}$ and $\hat{A} + \hat{B} + \hat{C} = \pi$, so that $\operatorname{sen}\hat{A} = \operatorname{sen}(\hat{B} + \hat{C})$ etc.

1) $\operatorname{sen}(\hat{B} + \hat{C})\operatorname{sen}(\hat{B} - \hat{C}) = \operatorname{sen}^2\hat{B} - \operatorname{sen}^2\hat{C}$. Thus, the first member is worth $2R[(\operatorname{sen}^2\hat{B} - \operatorname{sen}^2\hat{C}) + (\operatorname{sen}^2\hat{C} - \operatorname{sen}^2\hat{A}) + (\operatorname{sen}^2\hat{A} - \operatorname{sen}^2\hat{B})] = 0$.

2) $\operatorname{sen}(\hat{B} + \hat{C})\cos(\hat{B} - \hat{C}) = \frac{1}{2}(\operatorname{sen} 2\hat{B} + \operatorname{sen} 2\hat{C}) = \operatorname{sen}\hat{B}\cos\hat{B} + \operatorname{sen}\hat{C}\cos\hat{C}$; Multiplying by $2R$, we obtain the thesis.

3) Regrouping: $(b\cos\hat{C} + c\cos\hat{B}) + (c\cos\hat{A} + a\cos\hat{C}) + (a\cos\hat{B} + b\cos\hat{A}) = a + b + c$, by the projection theorem.

4) $\dfrac{a^2 - b^2}{c^2} = \dfrac{\operatorname{sen}^2\hat{A} - \operatorname{sen}^2\hat{B}}{\operatorname{sen}^2\hat{C}} = \dfrac{\operatorname{sen}(\hat{A} + \hat{B})\operatorname{sen}(\hat{A} - \hat{B})}{\operatorname{sen}^2(\hat{A} + \hat{B})} = \dfrac{\operatorname{sen}(\hat{A} - \hat{B})}{\operatorname{sen}(\hat{A} + \hat{B})}$.

5) According to the law of cosines, $c^2 + a^2 - b^2 = 2ac\cos\hat{B}$ and $c^2 + b^2 - a^2 = 2bc\cos\hat{A}$. The quotient is $\dfrac{a\cos\hat{B}}{b\cos\hat{A}} = \dfrac{\operatorname{sen}\hat{A}\cos\hat{B}}{\operatorname{sen}\hat{B}\cos\hat{A}} = \dfrac{\operatorname{tg}\hat{A}}{\operatorname{tg}\hat{B}}$.

6) $\dfrac{b + c}{a} = \dfrac{\operatorname{sen}\hat{B} + \operatorname{sen}\hat{C}}{\operatorname{sen}\hat{A}} = \dfrac{2\operatorname{sen}\frac{\hat{B} + \hat{C}}{2}\cos\frac{\hat{B} - \hat{C}}{2}}{2\operatorname{sen}\frac{\hat{A}}{2}\cos\frac{\hat{A}}{2}} = \dfrac{\cos\frac{\hat{B} - \hat{C}}{2}}{\operatorname{sen}\frac{\hat{A}}{2}}$, as $\operatorname{sen}\frac{\hat{B} + \hat{C}}{2} = \cos\frac{\hat{A}}{2}$.

7) Likewise, $\dfrac{b - c}{a} = \dfrac{2\cos\frac{\hat{B} + \hat{C}}{2}\operatorname{sen}\frac{\hat{B} - \hat{C}}{2}}{2\operatorname{sen}\frac{\hat{A}}{2}\cos\frac{\hat{A}}{2}} = \dfrac{\operatorname{sen}\frac{\hat{B} - \hat{C}}{2}}{\cos\frac{\hat{A}}{2}}$, as $\cos\frac{\hat{B} + \hat{C}}{2} = \operatorname{sen}\frac{\hat{A}}{2}$.

With $a = 4$, $b = 6$, $c = 8$: $p = 9$ and, by Heron’s formula, $S = \sqrt{9 \cdot 5 \cdot 3 \cdot 1} = 3\sqrt{15}$.

• Heights ($h = \frac{2S}{\text{lado}}$): $h_a = \frac{3\sqrt{15}}{2}$, $h_b = \sqrt{15}$, $h_c = \frac{3\sqrt{15}}{4}$.
• Medians ($m_a = \frac{1}{2}\sqrt{2b^2 + 2c^2 - a^2}$): $m_a = \sqrt{46}$, $m_b = \sqrt{31}$, $m_c = \sqrt{10}$.
• angle bisectors ($S_a = \frac{2}{b + c}\sqrt{bcp(p - a)}$): $S_a = \frac{12\sqrt{15}}{7}$, $S_b = 2\sqrt{6}$, $S_c = \frac{6\sqrt{6}}{5}$.
• Radius: $r = \frac{S}{p} = \frac{\sqrt{15}}{3}$ and $R = \frac{abc}{4S} = \frac{16\sqrt{15}}{15}$.

angle bisector divides the triangle into two, areas $\frac{1}{2}cS_a\operatorname{sen}\frac{\hat{A}}{2}$ and $\frac{1}{2}bS_a\operatorname{sen}\frac{\hat{A}}{2}$, whose sum is $\frac{1}{2}bc\operatorname{sen}\hat{A}$. Hence $S_a = \dfrac{2bc\cos\frac{\hat{A}}{2}}{b + c} = \dfrac{900\cos 15^\circ}{45} = 20\cos 15^\circ$.

Let us prove (1); the others are analogous. The circle previously inscribed at angle $\hat{A}$ touches the side $\overline{BC}$ and the extensions of $\overline{AB}$ and $\overline{AC}$, at points $T$, $T_c$ and $T_b$. The tangent segments drawn from a point are the same: $BT_c = BT$ and $CT_b = CT$. Then

and, like $AT_c = AT_b$, each measures $p$; in addition $BT_c = p - c$ and $CT_b = p - b$.

• The center $I_a$ is in angle bisector of $\hat{A}$, and in the right triangle $AT_cI_a$: $r_a = p\operatorname{tg}\frac{\hat{A}}{2}$.
• $I_a$ is at angle bisector of the exterior angle at $B$, which measures $180^\circ - \hat{B}$; in the right triangle $BT_cI_a$, the angle at $B$ is $90^\circ - \frac{\hat{B}}{2}$, therefore $r_a = (p - c)\operatorname{tg}\left(90^\circ - \frac{\hat{B}}{2}\right) = (p - c)\operatorname{cotg}\frac{\hat{B}}{2}$.
• Likewise, in $C$: $r_a = (p - b)\operatorname{cotg}\frac{\hat{C}}{2}$.

We use $r = \frac{S}{p}$ and $r_a = \frac{S}{p - a}$, $r_b = \frac{S}{p - b}$, $r_c = \frac{S}{p - c}$ (from $S = r_a(p - a)$, which comes out, as $S = pr$, from the triangle decomposition from ex-incenter). The hypothesis $r_a = r_b + r_c + r$ is

because $(p - b) + (p - c) = a$. Then $p(p - a) = (p - b)(p - c)$, that is, $p(b + c - a) = bc$. Multiplying by 2: $(a + b + c)(b + c - a) = 2bc$, or $(b + c)^2 - a^2 = 2bc$, or $b^2 + c^2 = a^2$. By exercise 372, the triangle is right-angled in $A$.

$R = \frac{abc}{4S}$ and $r = \frac{S}{p}$. So $R = 3r \iff abc \cdot p = 12S^2 = 12p(p - a)(p - b)(p - c) \iff abc = 12(p - a)(p - b)(p - c)$, and, like $p - a = \frac{b + c - a}{2}$ etc., $2abc = 3(b + c - a)(a + c - b)(a + b - c)$.

incenter $I$ is on angle bisector from $\hat{A}$ and is $r$ away from the $\overline{AB}$ side: $\alpha = AI = \dfrac{r}{\operatorname{sen}\frac{\hat{A}}{2}}$; similarly for $\beta$ and $\gamma$. By relation VII of exercise 410, $\operatorname{sen}\frac{\hat{A}}{2}\operatorname{sen}\frac{\hat{B}}{2}\operatorname{sen}\frac{\hat{C}}{2} = \frac{r}{4R}$; then

The printed statement has $\frac{abc}{2p}$ on the second member, which is wrong by a factor 2: on the equilateral triangle side 1, $\frac{\alpha\beta\gamma}{r} = \frac{2}{3}$, while $\frac{abc}{2p} = \frac{1}{3}$.

Let $R$ and $R'$ be the radii of the circumscribed circles. We have $\operatorname{sen}\hat{A}' = \operatorname{sen}\hat{A}$, $\operatorname{sen}\hat{B}' = \operatorname{sen}\hat{B}$ and $\hat{C}' = \pi - \hat{A}' - \hat{B}' = \hat{A} - \hat{B}$. By the law of sines,

As $\operatorname{sen}(\hat{A} + \hat{B})\operatorname{sen}(\hat{A} - \hat{B}) = \operatorname{sen}^2\hat{A} - \operatorname{sen}^2\hat{B}$, $bb' + cc' = 4RR'\operatorname{sen}^2\hat{A} = (2R\operatorname{sen}\hat{A})(2R'\operatorname{sen}\hat{A}') = aa'$ results.

We will use $a = 2R\operatorname{sen}\hat{A}$ (etc.), $S = \frac{1}{2}ab\operatorname{sen}\hat{C} = 2R^2\operatorname{sen}\hat{A}\operatorname{sen}\hat{B}\operatorname{sen}\hat{C}$, $S = pr$, $abc = 4RS$ and, from the figure of the inscribed circle, $\operatorname{tg}\frac{\hat{A}}{2} = \frac{r}{p - a}$ (etc.).

1) $\frac{1}{4}(a^2 \cdot 2\operatorname{sen}\hat{B}\cos\hat{B} + b^2 \cdot 2\operatorname{sen}\hat{A}\cos\hat{A}) = 2R^2\operatorname{sen}\hat{A}\operatorname{sen}\hat{B}(\operatorname{sen}\hat{A}\cos\hat{B} + \operatorname{sen}\hat{B}\cos\hat{A}) = 2R^2\operatorname{sen}\hat{A}\operatorname{sen}\hat{B}\operatorname{sen}\hat{C} = S$.

2) $a^2 - b^2 = 4R^2(\operatorname{sen}^2\hat{A} - \operatorname{sen}^2\hat{B}) = 4R^2\operatorname{sen}\hat{C}\operatorname{sen}(\hat{A} - \hat{B})$; substituting, the second member is worth $2R^2\operatorname{sen}\hat{A}\operatorname{sen}\hat{B}\operatorname{sen}\hat{C} = S$.

3) With $\operatorname{cotg}\frac{\hat{A}}{2} = \frac{p - a}{r}$ (etc.), the first member is worth $\frac{1}{r}[(b - c)(p - a) + (c - a)(p - b) + (a - b)(p - c)]$. The terms in $p$ add up to $p \cdot 0$ and the others add up to $-[(b - c)a + (c - a)b + (a - b)c] = 0$.

4) By the law of cosines, $\operatorname{cotg}\hat{A} = \frac{\cos\hat{A}}{\operatorname{sen}\hat{A}} = \frac{b^2 + c^2 - a^2}{4S}$ (etc.). Multiplying the first member by $4S$: $(b^2 - c^2)(b^2 + c^2 - a^2) + (c^2 - a^2)(c^2 + a^2 - b^2) + (a^2 - b^2)(a^2 + b^2 - c^2)$. The installments $b^4 - c^4 + c^4 - a^4 + a^4 - b^4$ add up to zero, and the remaining ones, $-a^2(b^2 - c^2) - b^2(c^2 - a^2) - c^2(a^2 - b^2)$, too.

5) The product is worth $\dfrac{r^3}{(p - a)(p - b)(p - c)} = \dfrac{r^3 p}{S^2} = \dfrac{r^3p}{p^2r^2} = \dfrac{r}{p}$.

6) With $q = ab + bc + ca$: expanding $(p - a)(p - b)(p - c) = pr^2$ and using $abc = 4Rrp$, we obtain $q = p^2 + r^2 + 4Rr$. So

7) $\operatorname{sen}^2\frac{\hat{A}}{2} = \frac{1 - \cos\hat{A}}{2} = \frac{a^2 - (b - c)^2}{4bc} = \frac{(p - b)(p - c)}{bc}$ (etc.). The product of the three squares is $\dfrac{[(p - a)(p - b)(p - c)]^2}{(abc)^2} = \dfrac{(pr^2)^2}{(4Rrp)^2} = \dfrac{r^2}{16R^2}$.

8) $\cos\hat{A} + \cos\hat{B} = 2\operatorname{sen}\frac{\hat{C}}{2}\cos\frac{\hat{A} - \hat{B}}{2}$ and $\cos\hat{C} = 1 - 2\operatorname{sen}^2\frac{\hat{C}}{2}$. As $\operatorname{sen}\frac{\hat{C}}{2} = \cos\frac{\hat{A} + \hat{B}}{2}$, the sum is worth $1 + 2\operatorname{sen}\frac{\hat{C}}{2}\left[\cos\frac{\hat{A} - \hat{B}}{2} - \cos\frac{\hat{A} + \hat{B}}{2}\right] = 1 + 4\operatorname{sen}\frac{\hat{A}}{2}\operatorname{sen}\frac{\hat{B}}{2}\operatorname{sen}\frac{\hat{C}}{2} = 1 + \frac{r}{R}$, for (7).

9) $a\cos\hat{A} + b\cos\hat{B} + c\cos\hat{C} = R(\operatorname{sen} 2\hat{A} + \operatorname{sen} 2\hat{B} + \operatorname{sen} 2\hat{C}) = 4R\operatorname{sen}\hat{A}\operatorname{sen}\hat{B}\operatorname{sen}\hat{C} = \frac{2S}{R} = \frac{2pr}{R}$, using $\operatorname{sen} 2\hat{A} + \operatorname{sen} 2\hat{B} = 2\operatorname{sen}\hat{C}\cos(\hat{A} - \hat{B})$ and $\operatorname{sen} 2\hat{C} = -2\operatorname{sen}\hat{C}\cos(\hat{A} + \hat{B})$.

As $S = 2R^2\operatorname{sen}\hat{A}\operatorname{sen}\hat{B}\operatorname{sen}\hat{C}$ and $\operatorname{sen}\hat{A} = \operatorname{sen}(\hat{B} + \hat{C})$, we get $R$; the sides come from the law of sines. The printed template writes $a = R\operatorname{sen}(\hat{B} + \hat{C})$ etc., without the factor 2: on the equilateral triangle side 1 this would give $a = \frac{1}{2}$.