In the interactive book: complete chapter X, with all demonstrations and exercises.

A bridge seen from the bank

Two points A and B are on the same bank of a river, separated by a known distance. A point C is on the other bank. By measuring the angles at A and B, we can calculate AC and BC without crossing. The triangle formed may not have any right angles.

Let’s call the angles $A,B,C$ and the opposite sides $a,b,c$. Always $A+B+C=180°$. The sides of a nondegenerate triangle are positive and each is less than the sum of the other two.

Demonstration of the law of cosines

Place A at the origin, B at $(c,0)$ and C at $(b\cos A,b\operatorname{sen} A)$. Side BC measures $a$. By the distance between two points, which comes from Pythagoras:

$$\begin{aligned} a^2&=(b\cos A-c)^2+(b\operatorname{sen} A)^2\\ &=b^2(\cos^2A+\operatorname{sen}^2A)+c^2-2bc\cos A\\ &=b^2+c^2-2bc\cos A. \end{aligned}$$

The expression also applies to $A$ obtuse, as the horizontal coordinate of C becomes negative. If $A=90°$, the cosine term is zero and we find Pythagoras again.

For example, for $b=5$, $c=7$, and $A=60°$, we have $a^2=25+49-70/2=39$, then $a=\sqrt{39}$.

Demonstration of the area formula

The height relative to the side $c$ measures $b\operatorname{sen} A$. If $A$ is obtuse, the height falls in the extension of the base and the sine remains positive. Therefore the area $K$ is:

$$K=\frac{bc\operatorname{sen} A}{2}.$$

Choosing other pairs of sides, we also obtain $K=ac\operatorname{sen} B/2=ab\operatorname{sen} C/2$. Equating the expressions and dividing by $abc/2$, which is positive, results:

$$\frac{\operatorname{sen} A}{a}=\frac{\operatorname{sen} B}{b}=\frac{\operatorname{sen} C}{c}.$$

Inverting, we arrive at the law of sines:

$$\frac a{\operatorname{sen} A}=\frac b{\operatorname{sen} B}=\frac c{\operatorname{sen} C}.$$

If $R$ is the radius of the circle that passes through the three vertices, the chord opposite $A$ measures $a=2R\operatorname{sen} A$. This results from dividing the triangle with two radii in half: for $A$ obtuse, we use the supplementary angle, with the same sine. So the common ratio above is $2R$, and $K=abc/(4R)$.

Solve without crossing the river

If $AB=c=20$ m and the measured angles are $A=45°$ and $B=60°$, then $C=75°$. By the law of sines:

$$\begin{aligned} b=AC&=\frac{20\operatorname{sen}60°}{\operatorname{sen}75°}\\ &=\frac{40\sqrt3}{\sqrt6+\sqrt2}\text{ m}. \end{aligned}$$

This is approximately 17,93 m. The $a=BC$ side is $20\operatorname{sen}45°/\operatorname{sen}75°$, approximately 14,64 m. The measures allow you to plan the crossing without entering the water.

What information is enough?

With two angles and one side, obtain the third angle and use the law of sines. With two sides and the angle between them, first use the law of cosines. With three valid sides, isolate a cosine and obtain the angle by the inverse function.

Two sides and an angle that is not between them require care. For example, if $A=30°$, $a=5$ and $b=8$, the law of sines gives $\operatorname{sen} B=4/5$. There are two choices: $B\approx53{,}13°$ or $B\approx126{,}87°$. Both leave a positive third angle. Therefore there are two possible triangles. Other data of this type may give one or no triangle.

Advanced workshop: special measures

The semiperimeter is $s=(a+b+c)/2$. The area can also be calculated using Heron’s formula:

$$K=\sqrt{s(s-a)(s-b)(s-c)}.$$

To demonstrate it, use $\cos A=(b^2+c^2-a^2)/(2bc)$ and $\operatorname{sen}^2A=1-\cos^2A$ in $K=bc\operatorname{sen} A/2$. Like this:

$$\begin{aligned} 16K^2&=4b^2c^2-(b^2+c^2-a^2)^2\\ &=((b+c)^2-a^2)(a^2-(b-c)^2)\\ &=(a+b+c)(b+c-a)\\ &\quad\cdot(a+b-c)(a-b+c)\\ &=16s(s-a)(s-b)(s-c). \end{aligned}$$

Since the area is positive, we take the positive root. The heights are $h_a=2K/a$, $h_b=2K/b$ and $h_c=2K/c$.

The inscribed circle touches the three sides. Its center is at the same distance $r$ from them. Dividing the triangle into three with height $r$, we have $K=(a+b+c)r/2=rs$, therefore $r=K/s$.

The median connects a vertex to the middle of the opposite side. Its measurement is:

$$m_a=\frac12\sqrt{2b^2+2c^2-a^2}.$$

To demonstrate, place the midpoint of the base at the origin, the extremes at $(-a/2,0)$ and $(a/2,0)$ and the vertex at $(u,v)$. Add the squared distance expressions: $b^2+c^2=2(u^2+v^2)+a^2/2$. As $u^2+v^2=m_a^2$, the formula results.

The inner angle bisector divides the angle in half. If its length is $t_a$, then:

$$t_a=\frac{2bc\cos(A/2)}{b+c}.$$

In fact, angle bisector divides the triangle into two areas $ct_a\operatorname{sen}(A/2)/2$ and $bt_a\operatorname{sen}(A/2)/2$. The sum is $bc\operatorname{sen} A/2$. Replacing $\operatorname{sen} A=2\operatorname{sen}(A/2)\cos(A/2)$ and canceling the positive factors, we obtain the expression.

You don’t need to memorize all of these tools at once. Choose a figure, identify what you know, and explain why the chosen formula applies.

In everyday life

  • Distance between boats. Two boats leaving the same port in different directions: the law of cosines gives the distance between them.
  • Land. Heron’s formula gives the area of ​​a triangular piece of land measuring only the three sides.
  • Robotic arms. The angle of the “elbow” of a robot or prosthesis is calculated using the law of cosines.
The distance between two boats leaving the same port, according to the law of cosines.

Exercises

Try solving it in your notebook before opening the solution. Exercises 1 and 2 are input; 3 and 4, for practice; 5 and 6, for deepening.

1. Two sides measure 6 and 8 cm and the angle between them is $60°$. Calculate the third side.

See resolution 1

$a^2=36+64-2\cdot6\cdot8\cdot1/2=52$. Logo $a=2\sqrt{13}$ cm.

2. Calculate the area of ​​a triangle with sides 8 and 10 cm forming $30°$.

See resolution 2

$K=8\cdot10\cdot\operatorname{sen}30°/2=20$ cm².

3. A triangle is $A=30°$, $B=45°$ and $a=6$ cm. Calculate $C$ and $b$.

See resolution 3

$C=180°-30°-45°=105°$. By the law of sines, $b=6(\sqrt2/2)/(1/2)=6\sqrt2$ cm.

4. A triangle has sides 5, 5 and 6 cm. Calculate area, height relative to the side 6 and inscribed radius.

See resolution 4

$s=8$ cm. By Heron, $K=\sqrt{8\cdot3\cdot3\cdot2}=12$ cm². The height is $2\cdot12/6=4$ cm. The inscribed radius is $12/8=1{,}5$ cm.

5. Calculate the median relative to the side of 6 cm in the triangle with sides 5, 5 and 6 cm.

See resolution 5

$m=\tfrac12\sqrt{2\cdot25+2\cdot25-36}=\tfrac12\sqrt{64}=4$ cm. In this isosceles triangle, this median is also height and angle bisector.

6. If $A=30°$, $a=4$ and $b=10$, is there a triangle?

See resolution 6

No. The law of sines would require $\operatorname{sen} B=b\operatorname{sen} A/a=10\cdot(1/2)/4=1{,}25$, impossible for a real angle.

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