Functions · Book or PDF

Functions 11

Chris Kirkpatrick, Barbara Alldred, Andrew Dmytriw, Shawn Godin, Angelo Lillo, David Pilmer, Susanne Trew, Noel Walker · Grade 11 · Study material generated by AI from this edition. Not affiliated with the publisher.

Chapter 5 · pp. 282–351

Chapter 5: Trigonometric Ratios

This chapter builds trigonometry from right triangles up to any triangle. It starts with the reciprocal ratios and the exact values for special angles, then extends the ratios to angles from 0° to 360° on the Cartesian plane. It finishes with trigonometric identities and the sine and cosine laws, which are used to solve triangles and real-world problems.

Reciprocal trigonometric ratios

The three primary ratios are sine, cosine and tangent. Each has a reciprocal ratio, found by dividing 1 by the primary ratio (p. 287):

$\csc\theta=\dfrac{1}{\sin\theta}=\dfrac{\text{hypotenuse}}{\text{opposite}}$
$\sec\theta=\dfrac{1}{\cos\theta}=\dfrac{\text{hypotenuse}}{\text{adjacent}}$
$\cot\theta=\dfrac{1}{\tan\theta}=\dfrac{\text{adjacent}}{\text{opposite}}$

The names are short for cosecant, secant and cotangent. Take a right triangle with opposite side 3, adjacent side 4 and hypotenuse 5. Then $\cot\theta=\frac{4}{3}$ , because adjacent is divided by opposite (pp. 287, 290).

Calculators. Most calculators have no cosecant, secant or cotangent key. Use the reciprocal of the matching primary key instead. For example, $\csc20^\circ=1\div\sin20^\circ$ , $\sec20^\circ=1\div\cos20^\circ$ and $\cot20^\circ=1\div\tan20^\circ$ . Make sure the calculator is in degree mode first (p. 287).

Why use a reciprocal ratio? Pick the ratio that puts the unknown side in the numerator. Then you solve with one multiplication. A clinometer measures an angle of elevation (above the horizontal) or an angle of depression (below the horizontal). Suppose a clinometer gives an elevation of 16.7° to the top of a tree about 3.0 m tall. The tree is the opposite side, and the distance MN along the ground is the adjacent side (p. 286).

Using tangent: $\tan16.7^\circ=\frac{3.0}{MN}$ . The unknown is on the bottom, so you must also divide by the tangent.
Using cotangent: $\cot16.7^\circ=\frac{MN}{3.0}$ , so $MN=3.0\cot16.7^\circ\approx10.0$ m.

Both methods give the same answer, but the cotangent equation is quicker to solve (pp. 286–287).

Comparing the six ratios

Take a right triangle with sides 3, 4 and 5, where angle A is opposite the side 3. Then:

$\sin A=0.60$ , $\cos A=0.80$ , $\tan A=0.75$
$\csc A\approx1.67$ , $\sec A=1.25$ , $\cot A\approx1.33$

Sine is the smallest primary ratio, and its reciprocal, cosecant, is the largest of all six. In general, the smallest primary ratio has the largest reciprocal (p. 288).

There is also a size rule. The hypotenuse is the longest side, so sine and cosine of an acute angle are at most 1. That makes cosecant and secant at least 1. Tangent can be less than, equal to or greater than 1, so cotangent can be too. At 45°, $\cot45^\circ=1$ (pp. 290–291).

Finding an angle from a reciprocal ratio

Calculators have no inverse secant key, so rewrite the ratio as its primary partner first. If $\sec\beta=2.69$ , then $\cos\beta=\frac{1}{2.69}$ . So $\beta=\cos^{-1}\left(\frac{1}{2.69}\right)\approx68^\circ$ . A primary ratio works directly: if $\tan\alpha=0.40$ , then $\alpha=\tan^{-1}(0.40)\approx22^\circ$ (p. 289).

Here is a full example. A right triangle has legs 8.0 cm and 20.0 cm. The hypotenuse is $\sqrt{8.0^2+20.0^2}=\sqrt{464}\approx21.5$ cm. One angle has $\tan\alpha=\frac{8.0}{20.0}=0.40$ , so it is about 22°. The other has $\sec\beta=\frac{21.5}{8.0}\approx2.69$ , so it is about 68° (p. 289).

Exact values for special angles

Some angles have ratios you can write exactly, using radicals instead of rounded decimals. They come from two "special triangles" (pp. 293–294).

The 45° triangle. A diagonal cuts a square of side 1 into two right isosceles triangles. The legs are 1, and the Pythagorean theorem gives a hypotenuse of $\sqrt2$ . So $\sin45^\circ=\cos45^\circ=\frac{1}{\sqrt2}$ . Multiply the top and bottom by $\sqrt2$ to get $\frac{\sqrt2}{2}\approx0.707$ . The legs are equal, so $\tan45^\circ=1$ .

The 30° and 60° triangle. Cut an equilateral triangle of side 2 in half along its height. The base splits into two parts of length 1. The right triangle has hypotenuse 2 and short leg 1. Its height satisfies $h^2=2^2-1^2=3$ , so $h=\sqrt3$ . The sides are 1, $\sqrt3$ and 2. The 30° angle is opposite the side 1, and the 60° angle is opposite $\sqrt3$ .

Reading the ratios from these sides (pp. 294, 296):

Angle	sin	cos	tan
30°	$\frac12$	$\frac{\sqrt3}{2}$	$\frac{\sqrt3}{3}$
45°	$\frac{\sqrt2}{2}$	$\frac{\sqrt2}{2}$	1
60°	$\frac{\sqrt3}{2}$	$\frac12$	$\sqrt3$

Whole-number denominators. Exact values are normally written without a radical in the denominator. Multiply the top and bottom by the radical: $\frac{1}{\sqrt3}=\frac{\sqrt3}{3}$ . This form also makes size estimates easy. Since $\sqrt2\approx1.4$ , half of it is about 0.7 (pp. 294–295).

Complementary angles. The side opposite one acute angle is adjacent to the other. So the sine of one acute angle equals the cosine of the other, and the two tangents are reciprocals. For example, $\sin30^\circ=\cos60^\circ=\frac12$ , $\cos30^\circ=\sin60^\circ=\frac{\sqrt3}{2}$ and $\tan30^\circ=\cot60^\circ$ . This is why the 30° and 60° rows use the same numbers (p. 295).

Spotting special angles. Tangent compares opposite to adjacent, and a larger angle has a longer opposite side. So tangent is below 1 from 0° to 45°, equals 1 at 45°, and is above 1 from 45° to 90°. You can also recognise special triangles from their sides (p. 296):

A right triangle with two equal sides has 45° angles.
A right triangle with one side half the hypotenuse has a 30° angle opposite that side.

Exact values of expressions

To evaluate an expression exactly, substitute the special-angle values, multiply, then add using a common denominator. For $(\sin45^\circ)(\cos45^\circ)+(\sin30^\circ)(\sin60^\circ)$ :

$\frac{\sqrt2}{2}\cdot\frac{\sqrt2}{2}+\frac12\cdot\frac{\sqrt3}{2}=\frac24+\frac{\sqrt3}{4}=\frac{2+\sqrt3}{4}$

(p. 295)

Watch the notation. Some students think $\tan^2 30^\circ$ means the tangent of a squared angle. It actually means the tangent value is squared: $\tan^230^\circ=(\tan30^\circ)^2=\left(\frac{\sqrt3}{3}\right)^2=\frac13$ . The angle stays 30° (p. 297).

Angles in standard position

To compare angles larger than 90°, we place them on the Cartesian plane. An angle is in standard position when its vertex is at the origin and its initial arm lies on the positive x-axis. The initial arm stays fixed. The terminal arm is the arm that rotates. A counterclockwise turn gives a positive angle. A 30° angle in standard position therefore has its terminal arm in quadrant 1. (pp. 299, 301)

The principal angle is the counterclockwise angle from the initial arm to the terminal arm, and it is always between 0° and 360°. When the terminal arm is in quadrant 2, 3 or 4, we also use the related acute angle. This is the acute angle between the terminal arm and the x-axis. Example: a terminal arm in quadrant 2 makes a 53° angle with the negative x-axis. Its principal angle is $180^\circ-53^\circ=127^\circ$ . (pp. 299, 301)

A negative angle is measured clockwise from the positive x-axis. A clockwise turn of 60° is $-60^\circ$ . Its terminal arm is in the same place as the arm of $360^\circ+(-60^\circ)=300^\circ$ . Two angles with the same terminal arm have the same trigonometric ratios, so $-60^\circ$ and $300^\circ$ match. (pp. 300–301)

How the quadrants relate to the related acute angle

Let $\theta$ be an acute angle. Reflect its triangle across the axes, and the new angles have ratios that equal those of $\theta$ or their negatives. The size stays the same. Only the sign can change, because reflecting flips the sign of $x$ , of $y$ , or of both.

Quadrant 2: $180^\circ-\theta$ (reflect in the y-axis; only $x$ becomes negative):

$\sin(180^\circ-\theta)=\sin\theta,\quad \cos(180^\circ-\theta)=-\cos\theta,\quad \tan(180^\circ-\theta)=-\tan\theta$

Example: $\sin150^\circ=\sin30^\circ=0.5$ , but $\cos150^\circ=-\frac{\sqrt3}{2}$ . (p. 301)

Quadrant 3: $180^\circ+\theta$ (reflect in both axes; $x$ and $y$ both become negative):

$\sin(180^\circ+\theta)=-\sin\theta,\quad \cos(180^\circ+\theta)=-\cos\theta,\quad \tan(180^\circ+\theta)=\tan\theta$

The tangent keeps its sign because $\frac yx$ is a negative divided by a negative. Example: $\tan225^\circ=\tan45^\circ=1$ . (pp. 301–302)

Quadrant 4: $360^\circ-\theta$ (reflect in the x-axis; only $y$ becomes negative):

$\sin(360^\circ-\theta)=-\sin\theta,\quad \cos(360^\circ-\theta)=\cos\theta,\quad \tan(360^\circ-\theta)=-\tan\theta$

Example: $\sin300^\circ=-\sin60^\circ=-\frac{\sqrt3}{2}$ . (pp. 301–302)

Ratios from a point on the terminal arm

Right-triangle words like "opposite" and "adjacent" do not work well for large angles. Instead, pick any point $P(x,y)$ on the terminal arm. Let $r$ be its distance from the origin. The coordinates are the legs of a right triangle, and $r$ is the hypotenuse, so

$r^2=x^2+y^2,\quad r>0.$

Example: for $P(3,4)$ , $r^2=9+16=25$ , so $r=5$ . The point $(-3,4)$ gives the same $r$ , because squaring removes the negative sign. The value of $r$ is always positive because it is a distance, which is also the radius of the circle through $P$ . (pp. 293–294, 298)

The six ratios are then:

$\sin\theta=\frac yr,\ \cos\theta=\frac xr,\ \tan\theta=\frac yx,\ \csc\theta=\frac ry,\ \sec\theta=\frac rx,\ \cot\theta=\frac xy$

These work for any angle from 0° to 360°, because $x$ and $y$ may be negative. Example: for $P(-3,4)$ , $r=5$ , so $\sin\theta=\frac45$ , $\cos\theta=-\frac35$ and $\tan\theta=-\frac43$ . (pp. 298, 303)

Finding the principal angle from a point. First find the related acute angle from the triangle's side lengths. Then use the quadrant to build the principal angle. For $P(-3,4)$ , use the side lengths $4$ and $5$ : $\sin\beta=\frac45$ , so $\beta\approx53^\circ$ . The point is in quadrant 2, so $\theta=180^\circ-53^\circ=127^\circ$ . For $P(3,4)$ in quadrant 1, the related angle is already the principal angle, about 53°. (pp. 293–295)

The angle 90°. Take $P(0,1)$ on the positive y-axis. No right triangle fits here, so use $x=0$ , $y=1$ , $r=1$ . Then $\sin90^\circ=\frac11=1$ and $\cos90^\circ=\frac01=0$ . The tangent is $\frac10$ , which is undefined because the denominator is zero. (p. 296)

Signs of the ratios: the CAST rule

Since $r$ is always positive, the signs of $x$ and $y$ decide the signs of the ratios. The CAST rule records this, starting in quadrant 1 and moving counterclockwise:

Quadrant 1: All three primary ratios are positive.
Quadrant 2: only Sine is positive ( $x<0$ , $y>0$ ).
Quadrant 3: only Tangent is positive (both $x$ and $y$ are negative, so $\frac yx>0$ ).
Quadrant 4: only Cosine is positive ( $x>0$ , $y<0$ ).

A reciprocal ratio has the same sign as its primary ratio. (pp. 298–299)

Finding every angle with a given ratio

To find all angles from 0° to 360° with a given ratio:

Use the sign of the ratio (CAST) to decide which two quadrants apply.
Ignore the sign and find the related acute angle.
Build the principal angle in each quadrant.

Example: $\csc\theta=-\frac{2\sqrt3}{3}$ . Take the reciprocal, so $\sin\theta=-\frac{\sqrt3}{2}$ . Sine is negative in quadrants 3 and 4. The related acute angle has $\sin=\frac{\sqrt3}{2}$ , so it is 60°. Then $\theta=180^\circ+60^\circ=240^\circ$ or $\theta=360^\circ-60^\circ=300^\circ$ . (pp. 297, 303)

A common mistake. Many students think the calculator's inverse key always gives the principal angle they need. It gives only one angle. That angle may be a related acute angle or a negative angle. For example, $\sin^{-1}\left(\frac45\right)\approx53^\circ$ , but the quadrant 2 angle with the same sine is 127°. Likewise, $\sin^{-1}(-0.8)\approx-53^\circ$ , which is a negative angle. The principal angles are about $180^\circ+53^\circ=233^\circ$ and $360^\circ-53^\circ=307^\circ$ . Always sketch the quadrant and check the sign before you decide on the answer. (pp. 295, 303, 311)

What a trigonometric identity is

A trigonometric identity is an equation with trigonometric ratios that is true for every angle where both sides are defined. This is different from an ordinary equation. The equation $\sin\theta=\tfrac12$ is true only for certain angles, but $\sin^2\theta+\cos^2\theta=1$ is true for every $\theta$ (pp. 315, 320).

If a side of the equation has a fraction, the denominator can never be zero. So every identity comes with restrictions, and you must state them. For example, $\tan\theta=\frac{\sin\theta}{\cos\theta}$ is an identity for all angles except those where $\cos\theta=0$ . In the range $0^\circ\le\theta\le360^\circ$ , those angles are $90^\circ$ and $270^\circ$ (pp. 315–316).

The identities to know

Some identities follow directly from definitions. Others are derived from relationships among the ratios (p. 319).

Reciprocal identities (these come from the definitions):

$\csc\theta=\frac{1}{\sin\theta},\qquad \sec\theta=\frac{1}{\cos\theta},\qquad \cot\theta=\frac{1}{\tan\theta}$

Each one needs its denominator to be nonzero. For instance, $\csc\theta$ needs $\sin\theta\ne0$ . At $30^\circ$ , $\sin30^\circ=\tfrac12$ , so $\csc30^\circ=2$ (p. 320).

Quotient identities:

$\tan\theta=\frac{\sin\theta}{\cos\theta}\ (\cos\theta\ne0),\qquad \cot\theta=\frac{\cos\theta}{\sin\theta}\ (\sin\theta\ne0)$

Pythagorean identities:

$\sin^2\theta+\cos^2\theta=1,\qquad 1+\tan^2\theta=\sec^2\theta,\qquad 1+\cot^2\theta=\csc^2\theta$

The second needs $\cos\theta\ne0$ . The third needs $\sin\theta\ne0$ (pp. 319–320).

Proving the basic identities with x, y and r

Pick a point $P(x,y)$ on the terminal arm of angle $\theta$ , with $r=\sqrt{x^2+y^2}$ . Then $\sin\theta=\frac yr$ , $\cos\theta=\frac xr$ and $\tan\theta=\frac yx$ . Writing every ratio this way is a good strategy when an identity has only a few ratios.

Quotient identity. The right side is

$\frac{\sin\theta}{\cos\theta}=\frac{y/r}{x/r}=\frac yr\cdot\frac rx=\frac yx,$

which is the left side, $\tan\theta$ . It fails only where $x=0$ , that is, where $\cos\theta=0$ (pp. 315–316).

Pythagorean identity. The left side is

$\frac{y^2}{r^2}+\frac{x^2}{r^2}=\frac{x^2+y^2}{r^2}.$

The Pythagorean theorem gives $r^2=x^2+y^2$ , so this becomes $\frac{r^2}{r^2}=1$ . A quick check at $30^\circ$ : $\left(\tfrac12\right)^2+\left(\tfrac{\sqrt3}{2}\right)^2=\tfrac14+\tfrac34=1$ (p. 316).

Another example is $\tan\theta\cos\theta=\frac yx\cdot\frac xr=\frac yr=\sin\theta$ (pp. 320–321).

Making new identities from known ones

You can build new identities by doing the same operation to both sides of an identity you already know. Start with $\sin^2\theta+\cos^2\theta=1$ :

Divide both sides by $\cos^2\theta$ . This gives $\tan^2\theta+1=\sec^2\theta$ .
Divide both sides by $\sin^2\theta$ . This gives $1+\cot^2\theta=\csc^2\theta$ .

The divisions are allowed only when $\cos\theta\ne0$ or $\sin\theta\ne0$ , which is where the restrictions come from (pp. 320–321, 336).

Rearranging the Pythagorean identity is also useful. Since $1-\sin^2x=\cos^2x$ , you can factor a difference of squares: $\cos^2x=(1-\sin x)(1+\sin x)$ . In the same way,

$\sin^4a-\cos^4a=(\sin^2a-\cos^2a)(\sin^2a+\cos^2a)=\sin^2a-\cos^2a.$

Spotting these patterns can shorten a proof a lot (pp. 320–321).

How to prove an identity

Write the left side (L.S.) and the right side (R.S.) separately.
Simplify the more complicated side until it matches the other, or change both sides to the same expression.
Do not move terms across the equal sign. That assumes the result is true, which is what you are trying to show.
State the restrictions, meaning the angles that make a denominator zero.

Tools for simplifying (pp. 319, 346):

Rewrite all ratios using $x$ , $y$ and $r$ .
Rewrite tangent and the reciprocal ratios in terms of sine and cosine.
Use a common denominator.
Factor.
Use the Pythagorean identity when you see $\sin^2\theta+\cos^2\theta$ .

For harder identities, a useful sequence is to switch to sine and cosine first, then combine fractions, then look for $\sin^2\theta+\cos^2\theta$ . For example,

$\tan\theta+\frac{1}{\tan\theta}=\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}=\frac{1}{\sin\theta\cos\theta}.$

Worked proofs

Proof by common denominator: $1+\cot^2\theta=\csc^2\theta$ . Use $\cot\theta=\frac{\cos\theta}{\sin\theta}$ and $\csc\theta=\frac1{\sin\theta}$ . The left side is

$1+\frac{\cos^2\theta}{\sin^2\theta}=\frac{\sin^2\theta}{\sin^2\theta}+\frac{\cos^2\theta}{\sin^2\theta}=\frac{\sin^2\theta+\cos^2\theta}{\sin^2\theta}=\frac{1}{\sin^2\theta}.$

The right side is $\left(\frac1{\sin\theta}\right)^2=\frac1{\sin^2\theta}$ , so the two sides match. Since $\sin\theta\ne0$ , the angles $0^\circ$ , $180^\circ$ and $360^\circ$ are excluded (p. 317).

Proof by factoring: $\tan\phi=\dfrac{\sin\phi+\sin^2\phi}{(\cos\phi)(1+\sin\phi)}$ . The right side is more complicated, so work on it. Factor $\sin\phi$ out of the numerator:

$\frac{\sin\phi(1+\sin\phi)}{(\cos\phi)(1+\sin\phi)}=\frac{\sin\phi}{\cos\phi}=\tan\phi.$

The common factor $1+\sin\phi$ cancels. The restriction is $\cos\phi\ne0$ (p. 318).

Proof of $\tan^2\theta+1=\sec^2\theta$ . The left side is

$\frac{\sin^2\theta}{\cos^2\theta}+1=\frac{\sin^2\theta+\cos^2\theta}{\cos^2\theta}=\frac{1}{\cos^2\theta}=\sec^2\theta,$

with the restriction $\cos\theta\ne0$ , so $\theta\ne90^\circ,270^\circ$ (pp. 320, 346).

Restrictions, and a common mistake

To find restrictions, look at every denominator in the original equation, including the ones hidden inside $\tan\theta$ , $\cot\theta$ , $\sec\theta$ and $\csc\theta$ . Find the angles that make each one zero, and exclude them. State these restrictions with the identity (pp. 321, 346).

A common wrong belief is that an equation must be an identity if both sides give the same value at one angle, such as $30^\circ$ . One matching value shows only that the equation is true for that angle. Many equations that are not identities agree at some angle. An identity must work for every allowed angle, so you need an algebraic proof. To show an equation is not an identity, one angle where the two sides differ is enough (p. 321).

Why a Graph Is Not a Proof

A graphing calculator can help you guess that an identity is true. Graph the left side as one function and the right side as another. If the two curves lie on top of each other, the sides might be equal. But this is only a hint. The screen shows a small window and has limited accuracy. Two graphs that differ by a tiny amount can look the same. A graph that is undefined at a single point can also look complete. Only an algebraic proof shows that the sides are equal for every angle where both are defined. (p. 322)

The Sine Law

Up to now the triangles have had a right angle. The sine law works in any triangle $ABC$ . It says that each side divided by the sine of its opposite angle gives the same number:

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

The flipped form is also true:

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

Choose the form that makes the unknown easy to isolate:

To find a side, put the sides in the numerators.
To find an angle, put the sines in the numerators. Then you can multiply both sides by a known side to get $\sin$ alone.

Example: a rope of length $a=5.9$ m and another of length $b=7.8$ m are given, with $\angle A=36^\circ$ . To find $\angle B$ , use $\frac{5.9}{\sin 36^\circ}=\frac{7.8}{\sin B}$ in the form $\frac{\sin B}{7.8}=\frac{\sin 36^\circ}{5.9}$ . (pp. 312, 327)

When you use a calculator, store the intermediate answers in memory. Round only at the very end. (p. 312)

When You Can Use the Sine Law

The sine law needs one complete pair: a known side and the angle opposite it. Two kinds of information give you this:

Two angles and any side (ASA or AAS). Find the third angle first, since the angles add to $180^\circ$ . Then you have a pair.
Two sides and an angle opposite one of them (SSA). The given angle and its opposite side form the pair.

With two sides and the contained angle (SAS), or with three sides (SSS), there is no side-and-opposite-angle pair to start from. The sine law cannot begin in those cases. (pp. 327, 335)

The Ambiguous Case (SSA)

The ambiguous case is the SSA situation. You know two sides and an angle that is opposite one of them, not between them. The given information may fit 0, 1 or 2 different triangles. (pp. 325, 327)

The number of triangles depends on the size of the given angle:

If the given angle is acute, 0, 1 or 2 triangles are possible.
If the given angle is obtuse, 0 or 1 triangle is possible.

Given angle $A$ is acute

Suppose $a$ , $b$ and $\angle A$ are known. Draw side $b$ along from $A$ . The height from the end of $b$ down to the base line is $h=b\sin A$ . Think of side $a$ as an arm that swings from the end of $b$ to reach the base line. Compare $a$ with $h$ and with $b$ :

$a<h$ : the arm is too short to reach the base. No triangle.
$a=h$ : the arm touches the base once, at a right angle. One right triangle.
$h<a<b$ : the arm can reach the base at two different points. Two triangles.
$a\ge b$ : only one point works. One triangle.

Example: let $A=36^\circ$ and $b=10.4$ cm. Then $h=10.4\sin36^\circ\approx6.1$ cm.

If $a<6.1$ cm, no triangle exists.
If $a=6.1$ cm, there is one right triangle.
If $a=7.0$ cm, then $6.1<7.0<10.4$ , so two triangles exist.
If $a\ge10.4$ cm, there is one triangle.

(pp. 327, 347)

Given angle $A$ is obtuse

The side opposite an obtuse angle must be the longest side of the triangle.

If $a>b$ , exactly one triangle exists.
If $a\le b$ , no triangle exists.

Example: an obtuse $A$ with $a=5$ and $b=8$ gives no triangle, because $a$ is not the longest side. (p. 327)

Because an obtuse given angle allows only one possible triangle, one sine law calculation is enough to find the other angle. (p. 327)

A Common Mistake: Trusting $\sin^{-1}$ Too Much

Some students think the inverse sine key gives the only possible angle in an SSA triangle. It does not. The calculator returns an angle from $0^\circ$ to $90^\circ$ . But an angle $B$ and its supplement $180^\circ-B$ have the same sine. If both angles fit inside a triangle, there are two answers. You must check for the second one. (pp. 313, 314, 325)

Solving an SSA Triangle with Two Solutions

Follow these steps:

Use the sine law to find the acute angle.
Find its supplement, $180^\circ-$ that angle.
Test each angle. It is valid only if it and the given angle add to less than $180^\circ$ .
Solve each triangle separately.

Example (the weather balloon ropes): $a=5.9$ m, $b=7.8$ m and $\angle A=36^\circ$ . The ropes' ground ends are the other two corners of the triangle. We want the side $c$ between them.

Step 1. Set up $\frac{\sin B}{7.8}=\frac{\sin 36^\circ}{5.9}$ . Then

$\sin B=\frac{7.8\sin36^\circ}{5.9}\quad\Rightarrow\quad \angle B\approx51^\circ$

Step 2. The supplement is $180^\circ-51^\circ=129^\circ$ .

Step 3. Check: $36^\circ+129^\circ=165^\circ<180^\circ$ , so both angles fit.

Step 4. Solve each triangle.

Triangle 1 ( $B\approx51^\circ$ ): $\angle C=180^\circ-(36^\circ+51^\circ)=93^\circ$ . Then $\frac{c}{\sin93^\circ}=\frac{5.9}{\sin36^\circ}$ gives $c\approx10.0$ m. This is the case where the two people stand on opposite sides of the balloon.
Triangle 2 ( $B=129^\circ$ ): $\angle C=180^\circ-(36^\circ+129^\circ)=15^\circ$ . Then $\frac{c}{\sin15^\circ}=\frac{5.9}{\sin36^\circ}$ gives $c\approx2.6$ m. This is the case where they stand on the same side.

The problem did not say where the people stood, so both answers are correct. Here $h=7.8\sin36^\circ\approx4.6$ m, and $5.9$ lies between $4.6$ and $7.8$ . That is exactly the two-triangle case. (pp. 312–315)

Bearings and the one-solution SSA case

A bearing tells you a direction. It is a clockwise angle measured from magnetic north. North is $0^\circ$ , east is $90^\circ$ , south is $180^\circ$ and west is $270^\circ$ . A bearing of $335^\circ$ points a little west of north. All north lines are parallel, so you can use alternate angles between parallel lines to move angles into a triangle. (p. 325)

A bearing problem can hide an SSA triangle. Karl's campsite is 15.6 m from a lake and 36.0 m from a lookout. The angle at the lake is $140^\circ$ . This angle is obtuse, so only one triangle is possible. A triangle cannot hold a second obtuse angle, so the sine law gives just one answer.

Put the angles on top of the sine law because you want an angle:

$\sin\theta=\frac{15.6\sin140^\circ}{36.0}\approx0.28,\quad \theta\approx16^\circ$

Alternate angles between parallel lines place this $16^\circ$ next to the north line. The bearing is therefore $90^\circ-16^\circ=74^\circ$ . (pp. 325–326)

Accuracy tip. Store intermediate values in your calculator's memory. Round only at the end. If you round an angle $B$ early, the side $c$ you find from it can be off in the last decimal place. (p. 312)

The cosine law

The sine law needs a side and its opposite angle. Sometimes you have no such pair. The cosine law covers those cases:

$a^2=b^2+c^2-2bc\cos A$

The matching forms are $b^2=a^2+c^2-2ac\cos B$ and $c^2=a^2+b^2-2ab\cos C$ . The angle in the cosine term is always the one opposite the side on the left. It works in any triangle. (pp. 322, 325)

Example (SAS). Let $b=26$ , $c=32$ and $A=60^\circ$ . Then $a^2=26^2+32^2-2(26)(32)(0.5)=868$ , so $a\approx29.5$ . (pp. 322, 325)

When to use it. Use the cosine law when you know two sides and the angle between them (SAS), or all three sides (SSS). In both cases there is no known side with its opposite angle, so the sine law cannot start. (pp. 325, 337)

Link to Pythagoras. If $A=90^\circ$ , then $\cos90^\circ=0$ . The term $-2bc\cos A$ disappears and you get $a^2=b^2+c^2$ . So the Pythagorean theorem is the cosine law for right triangles. The $-2bc\cos A$ term corrects for angles that are not $90^\circ$ . (pp. 323, 325)

Finding an angle from three sides. Solve for the cosine:

$\cos A=\frac{b^2+c^2-a^2}{2bc}$

Then use the inverse cosine. With $a=55$ , $b=26$ and $c=32$ cm:

$\cos A=\frac{26^2+32^2-55^2}{2(26)(32)}\approx-0.8,\quad A\approx143^\circ$

A negative cosine tells you the angle is obtuse. Be careful to pair each side with the correct angle. (pp. 323, 333)

Using the laws in applied problems

Ramp (auxiliary lines). A barn's cross-section is half a regular octagon with sides of 10 m. A 22.9 m ramp makes a $26^\circ$ angle with the ground. Its base is 15.6 m from the wall. Draw extra lines to make a triangle with two known sides and the angle between them. The base to the wall's foot is $15.6-5=10.6$ m. The cosine law gives the third side:

$b^2=10.6^2+22.9^2-2(10.6)(22.9)\cos26^\circ\approx200.4,\quad b\approx14.16\text{ m}$

The roof slope makes a $45^\circ$ - $45^\circ$ - $90^\circ$ triangle, so the part along the roof is $5\sqrt2\approx7.07$ m. The ramp reaches $14.16-7.07\approx7.09$ m up the slope. That leaves $10-7.09\approx2.9$ m to the flat roof. (pp. 321–322)

Both laws together. Mitchell's roof has sides of 8.0 m and 5.1 m enclosing $60^\circ$ . This is SAS, so use the cosine law first: $b^2=8.0^2+5.1^2-2(8.0)(5.1)\cos60^\circ=49.21$ , so $b\approx7.0$ m. Now you have a side with its opposite angle, so the sine law works:

$\sin\theta=\frac{5.1\sin60^\circ}{7.0},\quad\theta\approx39^\circ$

This is less than $40^\circ$ , so Mitchell can install the system himself. (pp. 333–334)

Three-dimensional problems

To solve a 3-D problem, follow these steps. (pp. 331, 337)

Sketch the solid.
Find unknown angles with geometry facts, such as angle sums in a triangle and parallel lines.
Redraw the flat triangles with the new information.
Use primary or reciprocal ratios in right triangles, and the sine and cosine laws in other triangles.

The flat triangles share sides. That shared side is the link between them.

Cliff height. In triangle $DBC$ the angles are $63^\circ$ and $42^\circ$ , so $C=180^\circ-105^\circ=75^\circ$ . Also $BD=68.5$ m. The sine law gives the shared side:

$BC=\frac{68.5\sin42^\circ}{\sin75^\circ}\approx47.45\text{ m}$

In right triangle $ABC$ , the height is $h=47.45\tan38^\circ\approx37.1$ m. (pp. 328–329)

Depression and bearings. An angle of depression is measured down from a horizontal line. That line is parallel to the ground, so alternate angles are equal. The depression angle from the observer equals the elevation angle from the object up to the observer. A difference of bearings gives the angle between two directions at the base. (pp. 330–331)

Two boats seen from a bridge. Suppose the bridge is 50 m high. The depression angles are $38^\circ$ and $35^\circ$ . The bearings are $230^\circ$ and $120^\circ$ , which differ by $110^\circ$ . Use tangent for each horizontal distance:

$b=\frac{50}{\tan38^\circ}\approx64.0\text{ m},\quad a=\frac{50}{\tan35^\circ}\approx71.4\text{ m}$

Then apply the cosine law with the $110^\circ$ angle:

$q^2=64.0^2+71.4^2-2(64.0)(71.4)\cos110^\circ,\quad q\approx111\text{ m}$

(pp. 330–331)

Parallax and relative error

Parallax is the apparent shift of an object when seen from two places. To find the Moon's distance, make a triangle from the Moon and two observers. The known side is the distance between the observers. The parallax angle is the angle at the Moon. The other two angles come from the $180^\circ$ angle sum. Then the sine law gives the Moon's distance.

Check your result against the known value with relative error:

$\text{relative error}=\frac{|\text{measured}-\text{actual}|}{\text{actual}}\times100\%$

Do this for three pairs of locations. The pair with the smallest relative error is the most accurate. (pp. 350–351)

Check yourself

Why does the cosine law reduce to the Pythagorean theorem when the angle is $90^\circ$ ?
A triangle has $a=7.2$ mm, $b=9.3$ mm and $A=35^\circ$ . How do you decide how many triangles are possible, and why is $\sin^{-1}$ alone not enough?
An angle in standard position has $\sin\theta<0$ and $\cos\theta>0$ . Which quadrant is it in, and how does CAST show this?
Show that $1+\cot^2\theta=\csc^2\theta$ by rewriting both sides in terms of sine and cosine. Which angles must be excluded?
The exact values for $30^\circ$ come from half of an equilateral triangle with side 2. How does that triangle give $\tan30^\circ$ , and why is $\tan30^\circ=\frac{\sqrt3}{3}$ written with a whole-number denominator?

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