How to find the shortest side of a triangle with angles

In the module Further trigonometry (Year 10), we introduced and proved the sine rule, which is used to find sides and angles in non-right-angled triangles.

In the triangle $ABC$, labelled as shown, we have

\[ \dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}. \]

Clearly, we may also write this as

\[ \dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c}. \]

In general, one of the three angles may be obtuse. The formula still holds true, although the geometric proof is slightly different.

Exercise 4

Find two expressions for $h$ in the diagram below, and hence deduce the sine rule.

Detailed description of diagram
Repeat the method of part (a), using the following diagram, to show that the sine rule holds in obtuse-angled triangles.

The triangle $ABC$ has $AB = 9\text{ cm}$, $\angle ABC = 76^\circ$ and $\angle ACB = 58^\circ$.

Find, correct to two decimal places,

Solution

Applying the sine rule gives \[ \dfrac{AC}{\sin 76^\circ} = \dfrac{9}{\sin 58^\circ} \] and so \begin{align*} AC = \dfrac{9 \sin 76^\circ}{\sin 58^\circ}\\ \approx 10.30\text{ cm} \qquad \text{(to two decimal places).} \end{align*}
To find $BC$, we first find the angle $\angle CAB$ opposite it. \begin{align*} \angle CAB = 180^\circ - 58^\circ - 76^\circ\\ = 46^\circ. \end{align*} Thus, by the sine rule, \[ \dfrac{BC}{\sin 46^\circ} = \dfrac{9}{\sin58^\circ} \] and so \[ BC \approx 7.63\text{ cm} \qquad \text{(to two decimal places).} \]

The ambiguous case

In the module Congruence (Year 8), it was emphasised that, when applying the SAS congruence test, the angle in question has to be the angle included between the two sides. For example, the following diagram shows two non-congruent triangles $ABC$ and $ABC'$ having two pairs of matching sides and sharing a common (non-included) angle.

Detailed description of diagram

Suppose we are told that a triangle $PQR$ has $PQ = 9$, $\angle PQR = 45^\circ$ and $PR = 7$. Then the angle opposite $PQ$ is not uniquely determined. There are two non-congruent triangles that satisfy the given data.

Detailed description of diagram

Applying the sine rule to the triangle, we have

\[ \dfrac{\sin \theta}{9} = \dfrac{\sin 45^\circ}{7} \] and so \begin{align*} \sin\theta = \dfrac{9\sin 45^\circ}{7}\\ \approx 0.9091. \end{align*}

Thus $\theta \approx 65^\circ$, assuming that $\theta$ is acute. But the supplementary angle is $\theta' \approx 115^\circ$. The triangle $PQR'$ also satisfies the given data. This situation is sometimes referred to as the ambiguous case.

Since the angle sum of a triangle is $180^\circ$, in some circumstances only one of the two angles calculated is geometrically valid.

The cosine rule

We know from the SAS congruence test that a triangle is completely determined if we are given two sides and the included angle. However, if we know two sides and the included angle in a triangle, the sine rule does not help us determine the remaining side or the remaining angles.

The second important formula for general triangles is the cosine rule.

Suppose $ABC$ is a triangle and that the angles $A$ and $C$ are acute. Drop a perpendicular from $B$ to the line interval $AC$ and mark the lengths as shown in the following diagram.

Detailed description of diagram

In the triangle $ABD$, applying Pythagoras' theorem gives

\[ c^2 = h^2 + (b - x)^2. \]

Also, in the triangle $BCD$, another application of Pythagoras' theorem gives

\[ h^2 = a^2 - x^2. \]

Substituting this expression for $h^2$ into the first equation and expanding,

\begin{align*} c^2 = a^2 - x^2 + (b - x)^2\\ = a^2 - x^2 + b^2 - 2bx + x^2\\ = a^2 + b^2 - 2bx. \end{align*}

Finally, from triangle $BCD$, we have $x = a\,\cos C$ and so

\[ c^2 = a^2 + b^2 - 2ab\,\cos C. \]

This last formula is known as the cosine rule.

Notice that, if $C = 90^\circ$, then since $\cos C = 0$ we obtain Pythagoras' theorem, and so we can regard the cosine rule as Pythagoras' theorem with a correction term.

The cosine rule is also true when the angle $C$ is obtuse. But note that, in this case, the final term in the formula will produce a positive number, because the cosine of an obtuse angle is negative. Some care must be taken in this instance.

By relabelling the sides and angles of the triangle, we can also write the cosine rule as $a^2 = b^2 + c^2 - 2bc\,\cos A$ and $b^2 = a^2 + c^2 - 2ac\,\cos B$.

Find the value of $x$ to one decimal place.

Solution

Applying the cosine rule gives

\begin{align*} x^2 = 7^2 + 8^2 - 2 \times 7 \times 8 \times \cos 110^\circ\\ = 113 + 112\cos70^\circ\\ \approx 151.306, \end{align*}

so $x \approx 12.3$ (to one decimal place).

Finding angles

If the three sides of a triangle are known, then the three angles are uniquely determined. (This is the SSS congruence test.) Again, the sine rule is of no help in finding the three angles, since it requires the knowledge of (at least) one angle, but we can use the cosine rule instead.

We can substitute the three side lengths $a$, $b$, $c$ into the formula $c^2 = a^2 + b^2 - 2ab\,\cos C$, where $C$ is the angle opposite the side $c$, and then rearrange to find $\cos C$ and hence $C$.

Alternatively, we can rearrange the formula to obtain

\[ \cos C = \dfrac{a^2+b^2-c^2}{2ab} \]

and then substitute. Students may choose to rearrange the cosine rule or to learn this further formula. Using this form of the cosine rule often reduces arithmetical errors.

Recall that, in any triangle $ABC$ labelled as shown, if $a < b$, then $\text{angle } A < \text{angle } B$.

A triangle has side lengths 6 cm, 8 cm and 11 cm. Find the smallest angle in the triangle.

Solution

The smallest angle in the triangle is opposite the smallest side.

Applying the cosine rule:

\begin{align*} 6^2 = 8^2 + 11^2 - 2 \times 8 \times 11 \times \cos \theta \\ \cos\theta = \dfrac{8^2+11^2-6^2}{2\times 8 \times 11}\\ =\dfrac{149}{176}. \end{align*}

So $\theta \approx 32.2^\circ$ (correct to one decimal place).

The area of a triangle

We saw in the module Introductory trigonometry (Years 9--10) that, if we take any triangle with two given sides $a$ and $b$ about a given (acute) angle $\theta$, then the area of the triangle is

\[ \text{Area} = \dfrac{1}{2} ab\,\sin\theta. \]

This formula also holds when $\theta$ is obtuse.

Exercise 5

A triangle has two sides of length 5 cm and 4 cm containing an angle $\theta$. Its area is 5 cm$^2$. Find the two possible (exact) values of $\theta$ and draw the two triangles that satisfy the given information.

Exercise 6

Write down two different expressions for the area of a triangle $ABC$ and derive the sine rule from them.

Next page - Content - Trigonometric identities

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In the $\triangle PQR$ angle P= 60 degrees & angle Q= 90 degrees . Find the length of the shortest side.

Any Ideas on how to begin

Many Thanks :)

From geometry we know that the sum of the angles in a triangle is 180°. Are there any relationships between the angles of a triangle and its sides?

First of all, you have probably observed that the longest side in a triangle is always opposite the largest angle, and the shortest side is opposite the smallest angle, as illustrated below.

In $\triangle FGH, \angle F=48\degree,$ and $\angle G$ is obtuse. Side $f$ is 6 feet long. What can you conclude about the other sides?

Solution.

Because $\angle G$ is greater than $90\degree\text{,}$ we know that $\angle F +\angle G$ is greater than $90\degree + 48\degree = 138\degree\text{,}$ so $\angle F$ is less than $180\degree-138\degree = 42\degree.$ Thus, $\angle H \lt \angle F \lt \angle G,$ and consequently $h \lt f \lt g\text{.}$ We can conclude that $h \lt 6$ feet long, and $g \gt 6$ feet long.

In isosceles triangle $\triangle RST\text{,}$ the vertex angle $\angle S = 72\degree\text{.}$ Which side is longer, $s$ or $t\text{?}$

It is also true that the sum of the lengths of any two sides of a triangle must be greater than the third side, or else the two sides will not meet to form a triangle. This fact is called the triangle inequality.

We cannot use the triangle inequality to find the exact lengths of the sides of a triangle, but we can find largest and smallest possible values for the length.

Solution.

We let $x$ represent the length of the third side of the triangle. By looking at each side in turn, we can apply the triangle inequality three different ways, to get

\begin{equation*} 7 \lt x+10, ~~~ 10 \lt x+7, ~~~ \text{and} ~~~ x \lt 10+7 \end{equation*}

We solve each of these inequalities to find

\begin{equation*} -3 \lt x, ~~~ 3 \lt x, ~~~ \text{and} ~~~ x \lt 17 \end{equation*}

We already know that $x \gt -3$ because $x$ must be positive, but the other two inequalities do give us new information. The third side must be greater than 3 inches but less than 17 inches long.

Can you make a triangle with three wooden sticks of lengths 14 feet, 26 feet, and 10 feet? Sketch a picture, and explain why or why not.

Answer.

No, $10+14$ is not greater than 26.

In Chapter 1 we used the Pythagorean theorem to derive the distance formula. We can also use the Pythagorean theorem to find one side of a right triangle if we know the other two sides.

A 25-foot ladder is placed against a wall so that its foot is 7 feet from the base of the wall. How far up the wall does the ladder reach?

Solution.

We make a sketch of the situation, as shown below, and label any known dimensions. We'll call the unknown height $h\text{.}$

\begin{align*} h^2 + 49 \amp = 625 \amp\amp \blert{\text{Subtract 49 from both sides.}}\\ h^2 \amp = 576 \amp\amp \blert{\text{Extract roots.}}\\ h \amp = \pm \sqrt{576} \amp\amp \blert{\text{Simplify the radical.}}\\ h \amp = \pm 24 \end{align*}

The height must be a positive number, so the solution $-24$ does not make sense for this problem. The ladder reaches 24 feet up the wall.

A baseball diamond is a square whose sides are 90 feet long. The catcher at home plate sees a runner on first trying to steal second base, and throws the ball to the second-baseman. Find the straight-line distance from home plate to second base.

Answer.

$90\sqrt{2} \approx 127.3$ feet

Keep in mind that the Pythagorean theorem is true only for right triangles, so the converse of the theorem is also true. In other words, if the sides of a triangle satisfy the relationship $a^2 + b^2 = c^2\text{,}$ then the triangle must be a right triangle. We can use this fact to test whether or not a given triangle has a right angle.

Delbert is paving a patio in his back yard, and would like to know if the corner at $C$ is a right angle.

Solution.

If is a right triangle, then its sides must satisfy $p^2 + q^2 = c^2\text{.}$ We find

\begin{align*} p^2 + q^2 \amp = 20^2 + 48^2 = 400 + 2304 = 2704\\ c^2 \amp = 52^2 = 2704 \end{align*}

Yes, because $p^2 + q^2 = c^2\text{,}$ the corner at $C$ is a right angle.

The sides of a triangle measure 15 inches, 25 inches, and 30 inches long. Is the triangle a right triangle?

The Pythagorean theorem relates the sides of right triangles. However, for information about the sides of other triangles, the best we can do (without trigonometry!) is the triangle inequality. Nor does the Pythagorean theorem help us find the angles in a triangle. In the next section we discover relationships between the angles and the sides of a right triangle.

Review the following skills you will need for this section.

$\displaystyle \:x \lt 3$
$\displaystyle \:x \le 8$
$\displaystyle \:x \le -1$
$\displaystyle \:x \gt -1$
Positive
Negative
Positive
Negative
Negative
Negative

Converse
Extraction of roots
Inequality

The longest side in a triangle is opposite the largest angle, and the shortest side is opposite the smallest angle.
Triangle Inequality: In any triangle, the sum of the lengths of any two sides is greater than the length of the third side.
Pythagorean Theorem: In a right triangle with hypotenuse $c,~~ a^2 +b^2 = c^2\text{.}$
If the sides of a triangle satisfy the relationship $~a^2 +b^2 = c^2~\text{,}$ then the triangle is a right triangle.

Is it always true that the hypotenuse is the longest side in a right triangle? Why or why not?
In $\triangle DEF\text{,}$ is it possible that $~d+e\gt f~$ and $~e+f\gt d~$ are both true? Explain your answer.
In a right triangle with hypotenuse $c\text{,}$ we know that $~a^2 +b^2 = c^2~\text{.}$ Is it also true that $~a + b = c~\text{?}$ Why or why not?
The two shorter sides of an obtuse triangle are 3 in and 4 in. What are the possible lengths for the third side?

Identify inconsistencies in figures #1-12
Use the triangle inequality to put bounds on the lengths of sides #13-16
Use the Pythagorean theorem to find the sides of a right triangle #17-26
Use the Pythagorean theorem to identify right triangles #27-32
Solve problems using the Pythagorean theorem #33-42

For Problems 1–12, explain why the measurements shown cannot be accurate.

If two sides of a triangle are 6 feet and 10 feet long, what are the largest and smallest possible values for the length of the third side?

Two adjacent sides of a parallelogram are 3 cm and 4 cm long. What are the largest and smallest possible values for the length of the diagonal?

If one of the equal sides of an isosceles triangle is 8 millimeters long, what are the largest and smallest possible values for the length of the base?

The town of Madison is 15 miles from Newton, and 20 miles from Lewis. What are the possible values for the distance from Lewis to Newton?

For Problems 17–22,

Make a sketch of the situation described, and label a right triangle.
Use the Pythagorean Theorem to solve each problem.

The size of a TV screen is the length of its diagonal. If the width of a 35-inch TV screen is 28 inches, what is its height?

If a 30-meter pine tree casts a shadow of 30 meters, how far is the tip of the shadow from the top of the tree?

The diagonal of a square is 12 inches long. How long is the side of the square?

The length of a rectangle is twice its width, and its diagonal is meters long. Find the dimensions of the rectangle.

For Problems 23–26, find the unknown side of the triangle.

For Problems 27–32, decide whether a triangle with the given sides is a right triangle.

9 in, 16 in, 25 in

12 m, 16 m, 20 m

5 m, 12 m, 13 m

5 ft, 8 ft, 13 ft

$5^2$ ft, $8^2$ ft, $13^2$ ft

$\sqrt{5}$ ft, $\sqrt{8}$ ft, $\sqrt{13}$ ft

Show that the triangle with vertices $(0,0)\text{,}$ $(6,0)$ and $(3,3)$ is an isosceles right triangle, that is, a right triangle with two sides of the same length.

Two opposite vertices of a square are $A(-9,-5)$ and $C(3,3)\text{.}$

Find the length of a diagonal of the square.
Find the length of the side of the square.

Find $\alpha, \beta$ and $h\text{.}$

Find $\alpha, \beta$ and $d\text{.}$

For Problems 41 and 42, make a sketch and solve.

The back of Brian's pickup truck is five feet wide and seven feet long. He wants to bring home a 9-foot length of copper pipe. Will it lie flat on the floor of the truck?
Find the length of the side of the square.

What is the longest curtain rod that will fit inside a box 60 inches long by 10 inches wide by 4 inches tall?

What are the coordinates of the other two vertices of the triangle? What is the length of the side joining those vertices?
Use the distance formula to compute the lengths of the other two sides of the triangle.
Show that the sides of the triangle satisfy the Pythagorean theorem, $a^2 + b^2 = c^2\text{.}$

There are many proofs of the Pythagorean theorem. Here is a simple visual argument.

What is the length of the side of the large square in the figure? Write an expression for its area.
Write another expression for the area of the large square by adding the areas of the four right triangles and the smaller central square.
Equate your two expressions for the area of the large square, and deduce the Pythagorean theorem.