International Baccalaureate · MYP 4

Pythagoras and Right-Angled Trigonometry

⌚ ~138 min ▢ 15 blocks ✓ 71 exercises ➣ Prerequisites : Pythagoras' Theorem, Similarity, Algebraic Expansion

I Pythagoras' Theorem

I.1 Proving the Converse and Finding Pythagorean Triples

Theorem — Pythagoras' Theorem

For any right-angled triangle with legs of length $\textcolor{colordef}{a}$ and $\textcolor{colorprop}{b}$ and a hypotenuse of length $\textcolor{olive}{c}$, the square of the hypotenuse is equal to the sum of the squares of the legs: $$ \textcolor{colordef}{a^2} + \textcolor{colorprop}{b^2} = \textcolor{olive}{c^2}. $$ This means the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the other two sides.

Theorem — Converse of Pythagoras' Theorem

If a triangle has side lengths $a$, $b$, and $c$, where $c$ is the longest side, and they satisfy $a^2 + b^2 = c^2$, then the triangle is a right-angled triangle, and the right angle is opposite the longest side, $c$.

Proof

Consider a triangle $ABC$ with $CB = a$, $CA = b$, $AB = c$, where $a^2 + b^2 = c^2$. We show that the angle at $C$ is right, using only Pythagoras' theorem (already proved).
Let $H$ be the foot of the perpendicular from $B$ to the line $\Line{CA}$, and let $x = CH$. If $H = C$, then $\Line{CB} \perp \Line{CA}$ and we are done; if $H = A$, then $BH = BA = c$ and $HA = 0$, so the relation $c^2 = BH^2 + HA^2$ used below holds directly (no triangle needed); suppose then $H \neq C$. Two cases, depending on which side of $C$ the foot $H$ falls.
Case 1: $H$ on the ray $\Ray{CA}$ (the angle at $C$ is acute). The triangles $BHC$ and $BHA$ are right-angled at $H$, so$$\begin{aligned}[t] c^2 &= BH^2 + HA^2 && \text{(Pythagoras' theorem in } \triangle BHA\text{)} \\ &= (a^2 - x^2) + (b - x)^2 && \text{(Pythagoras' theorem in } \triangle BHC\text{, and } HA = |b-x|\text{)} \\ &= a^2 + b^2 - 2bx. \end{aligned}$$

Case 2: $H$ on the opposite ray (the angle at $C$ is obtuse). Then $HA = b + x$, and the same two applications give$$c^2 = (a^2 - x^2) + (b + x)^2 = a^2 + b^2 + 2bx.$$

Now compare with the hypothesis $c^2 = a^2 + b^2$. In Case 2 we would have $c^2 = a^2+b^2+2bx > a^2 + b^2$ (since $x > 0$): impossible. So we are in Case 1, and $a^2+b^2-2bx = a^2+b^2$ forces $x = 0$, that is $H = C$: the line $\Line{CB}$ is perpendicular to $\Line{CA}$. The triangle is right-angled at $C$ — the vertex opposite the side $c$.
Bonus of this computation: $c^2 < a^2+b^2$ exactly when the angle at $C$ is acute, and $c^2 > a^2+b^2$ exactly when it is obtuse.

Definition — Pythagorean Triple

Three non-zero whole numbers $a$, $b$ and $c$ form a Pythagorean triple when $$a^2 + b^2 = c^2.$$ The triple is written $(a, b, c)$, with $c$ the largest number; $(a, b, c)$ and $(b, a, c)$ are the same triple. Examples: $(3, 4, 5)$, $(5, 12, 13)$, $(8, 15, 17)$, $(7, 24, 25)$. By the converse of Pythagoras' theorem, a triangle whose side lengths form a Pythagorean triple is right-angled. A triple whose three numbers have no common factor other than $1$ is called primitive: $(3, 4, 5)$ is primitive, $(6, 8, 10)$ is not.
About $3\,800$ years ago, a Babylonian clay tablet now called Plimpton~322 already recorded numbers from fifteen such triples.

Proposition — Multiples of a Pythagorean Triple

If $(a, b, c)$ is a Pythagorean triple and $k$ is a non-zero whole number, then $(ka, kb, kc)$ is also a Pythagorean triple. Conversely, dividing the three numbers of a triple by a common factor gives a Pythagorean triple; dividing them by their highest common factor gives a primitive triple, so every Pythagorean triple is a multiple of a primitive one.

Proof

Multiples. $ka$, $kb$ and $kc$ are non-zero whole numbers, and $$\begin{aligned}[t] (ka)^2 + (kb)^2 &= k^2a^2 + k^2b^2 && \text{(square of a product)}\\ &= k^2\left(a^2 + b^2\right) && \text{(factorising)}\\ &= k^2c^2 && (a^2 + b^2 = c^2)\\ &= (kc)^2. \end{aligned}$$ Division. If $d$ is a common factor, write $a = da'$, $b = db'$ and $c = dc'$ with whole numbers $a'$, $b'$, $c'$, non-zero like $a$, $b$, $c$. The same computation gives $d^2\left(a'^2 + b'^2\right) = d^2c'^2$; dividing by $d^2 \neq 0$, $a'^2 + b'^2 = c'^2$. When $d$ is the highest common factor, $a'$, $b'$ and $c'$ have no common factor other than $1$ (a common factor $e > 1$ would make $de$ a common factor of $a$, $b$, $c$ larger than $d$): the triple $(a', b', c')$ is primitive, and $(a, b, c)$ is $d$ times it.

Method — Recognising a Pythagorean Triple

To decide whether three numbers form a Pythagorean triple:

Check that they are non-zero whole numbers. If not, they do not form a triple — even when they are the sides of a right-angled triangle.
If the three numbers have a common factor, divide them by it: by the Proposition “Multiples of a Pythagorean Triple”, the smaller numbers form a triple exactly when the given ones do, and they are easier to square.
Call $c$ the largest number and compare $a^2 + b^2$ with $c^2$.

Example

Which of these groups of three numbers form a Pythagorean triple?

$29$, $20$ and $21$
$9$, $12$ and $16$
$45$, $60$ and $75$
$1.5$, $2$ and $2.5$

Answer

The largest number is $29$: $20^2 + 21^2 = 400 + 441 = 841$ and $29^2 = 841$, so $(20, 21, 29)$ is a Pythagorean triple (and a primitive one: a common factor of the three numbers would divide $20$ and $21$, which have no common factor other than $1$).
$9^2 + 12^2 = 81 + 144 = 225$ and $16^2 = 256$: $225 \neq 256$, so these numbers do not form a triple.
The three numbers are multiples of $15$: $45 = 15 \times 3$, $60 = 15 \times 4$, $75 = 15 \times 5$. Since $3^2 + 4^2 = 9 + 16 = 25 = 5^2$, $(3, 4, 5)$ is a triple, and so is $(45, 60, 75)$, $15$ times it.
$1.5$ and $2.5$ are not whole numbers: these numbers do not form a Pythagorean triple. Yet $1.5^2 + 2^2 = 2.25 + 4 = 6.25 = 2.5^2$: a triangle with these sides is right-angled — it is the $3$-$4$-$5$ triangle with every side divided by $2$.

Proposition — Generating Pythagorean Triples

For all whole numbers $m$ and $n$ with $m > n \geq 1$, the three numbers $$a = m^2 - n^2, \qquad b = 2mn, \qquad c = m^2 + n^2$$ form a Pythagorean triple. The formula does not give every Pythagorean triple — $(9, 12, 15)$, for example; multiplying the triples it gives by a non-zero whole number gives more of them. A construction equivalent to this formula already appears in Euclid's Elements, about $2\,300$ years ago.

Proof

Since $m > n \geq 1$, the three numbers are non-zero whole numbers ($m^2 > n^2$). Then $$\begin{aligned}[t] a^2 + b^2 &= \left(m^2 - n^2\right)^2 + (2mn)^2\\ &= m^4 - 2m^2n^2 + n^4 + 4m^2n^2 && \text{(perfect square, square of a product)}\\ &= m^4 + 2m^2n^2 + n^4\\ &= \left(m^2 + n^2\right)^2 && \text{(perfect square, read backwards)}\\ &= c^2. \end{aligned}$$

Example

Use the Proposition “Generating Pythagorean Triples” with $(m, n) = (2, 1)$, $(3, 2)$, $(4, 1)$ and $(3, 1)$. Which of the triples obtained are primitive?

Answer

For $(m, n) = (2, 1)$: $a = 2^2 - 1^2 = 3$, $b = 2 \times 2 \times 1 = 4$, $c = 2^2 + 1^2 = 5$. In the same way:

$m$	$n$	$m^2 - n^2$	$2mn$	$m^2 + n^2$	primitive?
$2$	$1$	$3$	$4$	$5$	yes
$3$	$2$	$5$	$12$	$13$	yes
$4$	$1$	$15$	$8$	$17$	yes
$3$	$1$	$8$	$6$	$10$	no

The first three are primitive: $3$ and $4$, $5$ and $12$, $15$ and $8$ have no common factor other than $1$, so neither have the three numbers. The last one, whose three numbers are even, is $(6, 8, 10) = 2 \times (3, 4, 5)$ once written in increasing order: the formula can give a multiple of a triple already found.

Skills to practice

Recognising Pythagorean Triples
Completing Pythagorean Triples

II Right-Angled Trigonometry

II.1 Sides of a Right-Angled Triangle

Discover

How tall is a tree you cannot climb? Stand $20\,\text{m}$ from its foot and, from ground level, measure the angle up to its top: $35^\circ$. You cannot put a tape to the top — but the height is already determined. The tree, the ground and your line of sight form a right-angled triangle, and every right-angled triangle with a $35^\circ$ angle is a scaled copy of this one (an enlargement or reduction of it), so the ratio $\dfrac{\text{height}}{\text{distance}}$ is always the same for $35^\circ$.

Those fixed ratios have names — sine, cosine, tangent — so that one side and one angle calculate the rest (here height $=20\tan 35^\circ$). The ratios work because all such triangles are enlargements or reductions of one another.

Answer

Once the trigonometric ratios are defined: the ratio $\dfrac{\text{height}}{\text{distance}}$ for $35^\circ$ is the tangent of $35^\circ$, and the calculator gives $\tan 35^\circ \approx 0.700$. So $$\text{height} = 20 \times \tan 35^\circ \approx 20 \times 0.700 = 14.0 \text{ m.}$$ With practice, this line becomes routine.

Definition — Right-Angled Triangle Sides

A right-angled triangle has one $90^\circ$ angle. We name the sides relative to a chosen acute angle $\theta$ (not the right angle).

The Hypotenuse (HYP) is the longest side, always opposite the right angle. It does not depend on the choice of $\theta$.
The Opposite (OPP) side is directly across from the angle $\theta$; it does not touch $\theta$.
The Adjacent (ADJ) side is the leg next to the angle $\theta$; it touches $\theta$ but is not the hypotenuse.

Example

In the triangle below, the angle $\theta$ is at vertex $B$. Identify the hypotenuse, the adjacent side, and the opposite side relative to angle $\theta$.

Answer

Relative to angle $\theta$ at $B$:

Hypotenuse: $\Segment{BC}$
Adjacent side: $\Segment{AB}$
Opposite side: $\Segment{AC}$

Skills to practice

Identifying the Sides of a Triangle

II.2 Trigonometric Ratios

The Foundation of Trigonometric Ratios

Why do trigonometric ratios work? Because two right-angled triangles with the same acute angle are an enlargement or a reduction of one another: all the lengths of one are the lengths of the other multiplied by the same scale factor $k$.
Consider any two right-angled triangles that share a common acute angle, $\theta$.

Both triangles have a right angle ($90^\circ$) and both share the angle $\theta$ — so their third angles are equal too (the angles of a triangle add up to $180^\circ$). Having the same angles, one triangle is an enlargement or a reduction of the other: every side of the second is the same multiple $k$ of the matching side of the first. This is exactly what Thales's theorem (the intercept theorem) guarantees: place the two angles $\theta$ on top of each other — the two opposite sides, both perpendicular to the common adjacent side, are parallel, so the two triangles are nested in a Thales configuration and their sides are proportional.

When we divide two sides of the same triangle, this factor $k$ cancels — the ratio comes out the same in both triangles. This means that for any right-angled triangle with the acute angle $\theta$: $$ \frac{\text{OPP}_1}{\text{HYP}_1} = \frac{\text{OPP}_2}{\text{HYP}_2}, \quad \frac{\text{ADJ}_1}{\text{HYP}_1} = \frac{\text{ADJ}_2}{\text{HYP}_2}, \quad \frac{\text{OPP}_1}{\text{ADJ}_1} = \frac{\text{OPP}_2}{\text{ADJ}_2}. $$ Since these ratios are constant for any given angle $\theta$, regardless of the size of the triangle, we can give them special names. These are the trigonometric ratios.

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Definition — The Three Trigonometric Ratios

For an acute angle $\theta$ in a right-angled triangle, we define the three main trigonometric ratios: sine, cosine, and tangent. $$ \sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}, \quad \cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}, \quad \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}. $$

A common mnemonic to remember these is SOH-CAH-TOA:

Sin = Opposite / Hypotenuse
Cos = Adjacent / Hypotenuse
Tan = Opposite / Adjacent

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Example

In the triangle below, find $\cos \theta$, $\sin \theta$, and $\tan \theta$.

Answer

Relative to $\theta$ at $B$:

Hypotenuse: $BC = 5$
Adjacent side: $AB = 4$
Opposite side: $AC = 3$

$$\begin{aligned} \cos \theta &= \frac{\text{ADJ}}{\text{HYP}} = \frac{4}{5}, \\ \sin \theta &= \frac{\text{OPP}}{\text{HYP}} = \frac{3}{5}, \\ \tan \theta &= \frac{\text{OPP}}{\text{ADJ}} = \frac{3}{4}. \end{aligned}$$

How Do the Ratios Vary?

Take the same ladder (a fixed hypotenuse) and lean it more steeply: the acute angle $\theta$ at its foot grows, the top climbs higher (the opposite side grows) and the foot comes closer to the wall (the adjacent side shrinks). So, as $\theta$ grows, $\sin\theta$ and $\tan\theta$ increase while $\cos\theta$ decreases.

Moreover, the opposite and the adjacent sides are both shorter than the hypotenuse: $\sin\theta$ and $\cos\theta$ are always between $0$ and $1$. The tangent, a ratio of the two legs, can exceed $1$.

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Proposition — Tangent Formula

For any acute angle $\theta$ of a right-angled triangle, $$\tan \theta = \frac{\sin \theta}{\cos \theta}$$

Proof

$$\begin{aligned} \tan \theta &= \frac{\text{OPP}}{\text{ADJ}}\\ &= \frac{\text{OPP}/\text{HYP}}{\text{ADJ}/\text{HYP}}\\ &= \frac{\sin \theta}{\cos \theta}.\\ \end{aligned}$$

Method — Using a Calculator

Trigonometric ratios for any acute angle can be calculated using a scientific calculator in degree mode. Always check that the calculator is set to “DEG” (degrees) before calculating.

Example

In the triangle below, find $x$, rounded to one decimal place.

Answer

Relative to the angle $30^\circ$ at $B$, the side of length $3$ is the hypotenuse and the side $x$ is adjacent to the angle. $$\begin{aligned} \cos \theta &= \frac{\text{ADJ}}{\text{HYP}} \\ \cos(30^\circ) &= \frac{x}{3} \\ x &= 3 \times \cos(30^\circ) \\ x &\approx 3 \times 0.866 \\ x &\approx 2.6 \quad \text{(rounded to one decimal place)} \end{aligned}$$

Example

In the triangle below, find $x$, rounded to one decimal place.

Answer

With respect to the $40^\circ$ angle at $B$: the side $AC = 5$ is opposite, the side $AB = x$ is adjacent. The ratio that links the opposite and the adjacent sides is the tangent: $$\begin{aligned} \tan(40^\circ) &= \frac{5}{x} \\ x \times \tan(40^\circ) &= 5 \\ x &= \frac{5}{\tan(40^\circ)} \\ x &\approx \frac{5}{0.839} \approx 6.0 \text{ cm} \quad \text{(rounded to one decimal place)} \end{aligned}$$ The unknown is in the denominator: multiply both sides by $x$, then divide by $\tan(40^\circ)$ — the unknown is never left under the fraction bar.

Skills to practice

Calculating Trigonometric Ratios
Calculating Side Lengths

II.3 Finding Angles

Trigonometric ratios can be used to find unknown angles in right-angled triangles when at least two side lengths are known.

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Definition — Inverse Trigonometric Functions

In a right-angled triangle with an acute angle $\theta$: $$ \theta = \cos^{-1}\left(\frac{\text{ADJ}}{\text{HYP}}\right), \quad \theta = \sin^{-1}\left(\frac{\text{OPP}}{\text{HYP}}\right), \quad \theta = \tan^{-1}\left(\frac{\text{OPP}}{\text{ADJ}}\right). $$ The notation $\cos^{-1}$ (read “inverse cosine”) is the calculator function that recovers the angle from the ratio — it does not mean $\frac{1}{\cos\theta}$.

Example

In the triangle below, find the angle $\theta$.

Answer

We know the lengths of the adjacent side ($AB = 0.5$) and the hypotenuse ($BC = 1$) relative to $\theta$. We can use the inverse cosine function: $$\begin{aligned} \theta &= \cos^{-1}\left(\frac{\text{ADJ}}{\text{HYP}}\right) \\ &= \cos^{-1}\left(\frac{0.5}{1}\right) \\ &= 60^\circ. \end{aligned}$$

Example

A slide is $4$ m long and its top is $2.4$ m above the ground. Find the angle the slide makes with the ground, rounded to the nearest degree.

Answer

The slide, the ground and the vertical line from the top form a right-angled triangle. With respect to the wanted angle $\theta$ (at the foot of the slide), we know the opposite side ($2.4$ m, the height) and the hypotenuse ($4$ m, the slide): the ratio that links them is the sine, so the inverse function to use is $\sin^{-1}$. $$\begin{aligned} \sin\theta &= \frac{\text{OPP}}{\text{HYP}} = \frac{2.4}{4} = 0.6 \\ \theta &= \sin^{-1}(0.6) \approx 36.9^\circ \approx 37^\circ. \end{aligned}$$ The choice of the inverse function follows the two known sides: opposite and hypotenuse $\to \sin^{-1}$; adjacent and hypotenuse $\to \cos^{-1}$; opposite and adjacent $\to \tan^{-1}$.

Skills to practice

Calculating Angles

II.4 Solving Problems with Trigonometry

Trigonometric ratios are powerful tools for solving a wide range of problems involving right-angled triangles, especially in real-world contexts. To solve these problems effectively, follow the structured steps below:

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Definition — Angle of Elevation and Angle of Depression

When an observer looks at an object, the angle of elevation is the angle between the horizontal and the line of sight when the object is above the observer; the angle of depression is the angle between the horizontal and the line of sight when the object is below the observer.

From the top $P$ of the cliff, the boat $B$ is seen under the angle of depression $\alpha$; from the boat, the top $P$ is seen under the angle of elevation $\alpha$. The two horizontals are parallel, so these two angles are equal (alternate angles). For the calculation, use the angle of elevation at $B$: it is an angle of the right-angled triangle $PFB$; the angle of depression, at $P$, lies outside the triangle (the triangle's angle at $P$ is $90^\circ - \alpha$).

Method — Solving Real-World Trigonometry Problems

Draw a clear diagram representing the situation described in the problem.
Label the unknown (side or angle) to be found. Use $x$ for a side and $\theta$ for an angle if possible.
Identify a right-angled triangle within the diagram.
Write an equation relating an angle and two sides of the triangle using the appropriate trigonometric ratio.
Solve the equation to find the unknown value.
State the answer clearly, including appropriate units, in a complete sentence.

Example

The string of a kite, taut and straight, is $50$ m long; its lower end is held at ground level, and the string makes a $40^\circ$ angle with the horizontal ground. How high is the kite flying? Round to the nearest metre.

Answer

We follow the six steps of the method.

Diagram: the string, the ground and the vertical line through the kite.
Unknown: the height $h$ of the kite.
Right-angled triangle: the right angle is between the ground and the vertical; the $40^\circ$ angle is at the hand, on the ground: it is the angle of elevation of the kite.
Equation: with respect to the $40^\circ$ angle, $h$ is the opposite side and the string ($50$ m) is the hypotenuse, so the sine is the ratio to use: $\sin(40^\circ) = \dfrac{h}{50}$.
Solve: $h = 50 \times \sin(40^\circ) \approx 50 \times 0.643 \approx 32.1$.
Answer: the kite is flying about $32$ m above the ground.

Skills to practice

Solving Real-World Trigonometry Problems
Writing the Solution

III Bearings

III.1 True Bearings

Trigonometry is a powerful tool for navigation and surveying. To describe the direction of travel, we use a standardised system called bearings.

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Definition — True Bearings

A true bearing is the direction to an object from a point, measured as an angle in a clockwise direction from the North line. The angle is measured in degrees from $000^\circ$ up to, but not including, $360^\circ$ (a full turn is written $000^\circ$), and bearings are always written using three figures (for example, $000^\circ$, $040^\circ$, $135^\circ$, $270^\circ$).

Example

The bearing of A from P is $135^\circ$.
The bearing of B from P is $040^\circ$.

Method — Key Geometric Principles for Bearings

Solving bearing problems relies on two key geometric principles that arise from the fact that all North lines on a map or diagram are parallel.

Parallel Line Angles: Because North lines are parallel, angle properties (such as alternate interior or co-interior angles) transfer a direction from one point to another. This is essential for finding the angles of the right-angled triangle formed by the path and the North–South line.
Back Bearings: The bearing from point A to point B is the forward bearing. The bearing from point B back to point A is the back bearing. A back bearing always differs by $180^\circ$ from the forward bearing (add or subtract $180^\circ$ and, if necessary, bring the result back between $000^\circ$ and $360^\circ$).

Skills to practice

Determining Bearings from a Diagram
Finding Back Bearings
Drawing and Explaining Bearings

III.2 Finding North and East Components

Method — Finding the North and East Components of a Journey

Unless it runs due North, East, South or West, a journey of length $d$ on a bearing $\theta$ is the hypotenuse of a right-angled triangle whose two other sides are its North–South part and its East–West part.

Draw the North–South line at the starting point, the journey, and the East–West segment from the end of the journey to the North–South line: these three segments form the right-angled triangle.
Find the acute angle $\alpha$ between the journey and the North–South line: $\alpha = \theta$ for a bearing between $000^\circ$ and $090^\circ$, $\alpha = 180^\circ - \theta$ between $090^\circ$ and $180^\circ$, $\alpha = \theta - 180^\circ$ between $180^\circ$ and $270^\circ$, $\alpha = 360^\circ - \theta$ between $270^\circ$ and $360^\circ$.
Compute the two parts in the right-angled triangle: the North–South part is $d\cos\alpha$ (the side adjacent to $\alpha$), the East–West part is $d\sin\alpha$ (the side opposite $\alpha$).
State the directions: north for a bearing below $090^\circ$ or above $270^\circ$, south for a bearing between $090^\circ$ and $270^\circ$; east for a bearing between $000^\circ$ and $180^\circ$, west for a bearing between $180^\circ$ and $360^\circ$.

A bearing of $000^\circ$, $090^\circ$, $180^\circ$ or $270^\circ$ points along the North–South or the East–West line: there is no triangle, one part is $d$ and the other is $0$.

Example

A boat sails $15$ km from a harbour $H$ on a bearing of $125^\circ$. How far south and how far east of $H$ is it (to $0.1$ km)?

Answer

The bearing lies between $090^\circ$ and $180^\circ$: the boat is south and east of $H$, and the journey makes the angle $\alpha = 180^\circ - 125^\circ = 55^\circ$ with the South line. $$\begin{aligned}[t] \text{south part} &= 15\cos 55^\circ \approx 8.6\ \text{km} && {\scriptstyle(\text{adjacent side})}\\ \text{east part} &= 15\sin 55^\circ \approx 12.3\ \text{km} && {\scriptstyle(\text{opposite side})} \end{aligned}$$ The boat is about $\mathbf{8.6}$ km south and $\mathbf{12.3}$ km east of the harbour. Check: $8.6^2 + 12.3^2 \approx 225 = 15^2$ (Pythagoras).

Skills to practice

Calculating North and East Components

IV Going further

Skills to practice

Investigating Pythagorean Triples
Solving Problems with Two Right-Angled Triangles
Reasoning and Solving Exam-Style Problems
Quiz

\(m\)	\(n\)	\(m^2 - n^2\)	\(2mn\)	\(m^2 + n^2\)	primitive?
\(2\)	\(1\)	\(3\)	\(4\)	\(5\)	yes
\(3\)	\(2\)	\(5\)	\(12\)	\(13\)	yes
\(4\)	\(1\)	\(15\)	\(8\)	\(17\)	yes
\(3\)	\(1\)	\(8\)	\(6\)	\(10\)	no