Right triangles and trigonometry

Look for the small square that marks a right angle. With no figure, the words can still give you one, like a ladder leaning on a wall or a walk due east and then due north.

The last line usually reads like one of these:

• “Which of the following is closest to the length of the triangle's hypotenuse?”
• “In the triangle shown, what is the value of $\tan x^\circ$?”
• “What is the value of $\sin A$?”

The two sides that meet at the right angle are the legs. The hypotenuse is across from the right angle, and it's always the longest side. If there's no figure, draw one on scratch paper and write every number on it.

A figure is drawn to scale unless it says “Note: Figure not drawn to scale.” When a figure has that note, use only the numbers.

It's on the reference sheetThe page of formulas you can open at any point in the Math section.Read the lesson. Say the legs are 5 and 8.

1. Square each leg and add.

   $$c^2=5^2+8^2$$

2. Work out the squares.

   $$c^2=25+64=89$$

3. Take the square root last.

   $$c=\sqrt{89}\approx9.43$$

To find a leg, subtract the leg's square from the hypotenuse's square. Say the hypotenuse is 75 and one leg is 21.

1. The hypotenuse's square comes first.

   $$b^2=75^2-21^2$$

2. Work out the squares.

   $$b^2=5625-441=5184$$

3. $72^2=5184$.

   $$b=72$$

Some choices are equationsTwo expressions set equal to each other, like 5x + 7 = 22.Read the lesson. If the legs are 7 and $n$ and the hypotenuse is 12, the equation is $7^2+n^2=12^2$. A choice like $7^2+12^2=n^2$ treats 12 as a leg.

For a “closest to” question, type sqrt in Desmos, then 5^2+8^2. It shows about 9.434.

Square a side like $2\sqrt6$ one piece at a time. Say the legs are $2\sqrt6$ and 5.

1. Square the 2 and square the root.

   $$(2\sqrt6)^2=2^2\cdot(\sqrt6)^2$$

2. Squaring a square root just removes it.

   $$=4\cdot6=24$$

3. Add the other leg's square, 25.

   $$c^2=24+5^2=49$$

4. $7^2=49$.

   $$c=7$$

Some choices write a length as a number times a square root. Pull the biggest square number out from under the root. Legs 6 and 12 give $\sqrt{36+144}=\sqrt{180}$, and $180=36\cdot5$, so the hypotenuse is $6\sqrt5$.

A diagonal cuts a rectangle into two right triangles. The rectangle's sides are the legs, and the diagonal is the hypotenuse.

If a rectangle's corners all sit on a circle, its diagonal is a diameterThe distance straight across a circle through the center, which is twice the radius.Read the lesson. A rectangle 9 by 12 has diagonal 15, so that circle's radiusThe distance from the center of a circle out to its edge.Read the lesson is 7.5.

When one side is described using the other, call the shorter side $x$. A rectangular garden has a diagonal path 26 meters long, and its length is 14 meters more than its width. What's the garden's area?

1. Width $x$, length $x+14$, diagonal 26.

   $$x^2+(x+14)^2=26^2$$

2. Multiply out $(x+14)(x+14)$. And $26^2=676$.

   $$x^2+x^2+28x+196=676$$

3. Add the $x^2$ terms and subtract 676 from both sides.

   $$2x^2+28x-480=0$$

4. Divide every term by 2.

   $$x^2+14x-240=0$$

5. 24 and $-10$ multiply to $-240$ and add to 14.

   $$(x+24)(x-10)=0$$

6. A width can't be $-24$.

   $$x=10$$

7. The length is $10+14=24$. Multiply for the area.

   $$10\cdot24=240$$

The area is 240 square meters.

In Desmos, type x^2+(x+14)^2=676. It draws a vertical line at each solution. Click where the right-hand line crosses the horizontal axisOne of the two lines a graph is measured against, one running across and one running up. to read 10.

Now try one where the diagonal has a square root in it.

Stand at the angle the question names. The leg across from it is the opposite side, and the leg touching it is the adjacent side. The hypotenuse doesn't change.

SOH CAH TOA is how you remember it. Sine is Opposite over Hypotenuse, Cosine is Adjacent over Hypotenuse, and Tangent is Opposite over Adjacent.

Here the right angle is at $M$, $LM=16$, $MN=12$ and $LN=20$. Standing at $L$, $MN$ is opposite and $LM$ is adjacent. So $\sin L=\frac{12}{20}$ and $\cos L=\frac{16}{20}$.

Standing at $N$, the legs trade jobs. $LM$ is now opposite, so $\tan N=\frac{16}{12}$.

Don't reduce a ratioA comparison of two amounts, like 3 to 5.Read the lesson unless the choices do. If a question gives $\sin L=\frac{12}{c}$ and asks for the hypotenuse $c$, it's 20. Reducing to $\frac35$ first would give 5.

Say a right triangle has legs 20 and 99, and the question asks for $\sin x^\circ$, where $x^\circ$ is the angle across from the 20.

1. Sine needs the hypotenuse, so find it.

   $$c^2=20^2+99^2=400+9801=10201$$

2. $101^2=10201$.

   $$c=101$$

3. Opposite over hypotenuse.

   $$\sin x^\circ=\frac{20}{101}$$

$\frac{20}{99}$ is the tangent, and $\frac{99}{101}$ is the cosine. You can answer from the sides without ever finding the angle.

This next one gives you the hypotenuse instead, so you subtract to find a leg first.

+−100%Reset⤢

Blessing records the triangular route shown in her log.

In the triangle shown, what is the value of $\tan x^{\circ}$?

1. A

   $\frac{5}{12}$

2. B

   $\frac{\sqrt{194}}{5}$

3. C

   $\frac{13}{5}$

4. D

   $\frac{12}{5}$

Show the answer

• A

  $\frac{5}{12}$

  This reverses the tangent ratioA comparison of two amounts, like 3 to 5.Read the lesson by dividing the adjacent leg, $15$, by the opposite leg, $36$, giving $\frac{5}{12}$.

• B

  $\frac{\sqrt{194}}{5}$

  This treats the hypotenuseThe longest side of a right triangle, the one opposite the right angle.Read the lesson, $39$, as a leg and calculates the missing side using $\sqrt{39^2+15^2}=3\sqrt{194}$. Dividing that result by $15$ gives $\frac{\sqrt{194}}{5}$.

• C

  $\frac{13}{5}$

  This treats the hypotenuseThe longest side of a right triangle, the one opposite the right angle.Read the lesson, $39$, as the side opposite angle $x$, giving $\frac{39}{15}=\frac{13}{5}$.

• D

  $\frac{12}{5}$

  Correct answer

  Use the Pythagorean theorem to find the opposite leg: $\sqrt{39^2-15^2}=\sqrt{1296}=36$. Therefore, $\tan x^\circ=\frac{\text{opposite}}{\text{adjacent}}=\frac{36}{15}=\frac{12}{5}$.

Solution

The side adjacent to angle $x$ is $15$, and the hypotenuseThe longest side of a right triangle, the one opposite the right angle.Read the lesson is $39$. Let the opposite leg be $r$. The Pythagorean theorem gives $r^2+15^2=39^2$, so $r^2=1521-225=1296$ and $r=36$. Tangent is the ratioA comparison of two amounts, like 3 to 5.Read the lesson of the opposite leg to the adjacent leg. Thus, $\tan x^\circ=\frac{36}{15}=\frac{12}{5}$.

Report a problem

These two side patterns are on the reference sheet.

A 45-45-90 triangle is half a square. With legs of 7, the hypotenuse is $7\sqrt2$ and the perimeterThe distance all the way around a flat shape.Read the lesson is $14+7\sqrt2$. Every 45-45-90 perimeter has two parts like that, one equal to the other times $\sqrt2$, and the smaller part is the hypotenuse. Here that's $7\sqrt2$, about 9.9.

A 30-60-90 triangle is half an equilateral triangle. The short leg $x$ is across from $30^\circ$, the long leg $x\sqrt3$ is across from $60^\circ$, and the hypotenuse is $2x$.

An equilateral triangle's height splits it into two of these. With sides of 10, each half has hypotenuse 10 and short leg 5, so the height is $5\sqrt3$.

Use SOH CAH TOA on these sides to get the values.

The tangents are $\tan30^\circ=\frac{\sqrt3}{3}$, $\tan45^\circ=1$, and $\tan60^\circ=\sqrt3$.

To check one in Desmos, switch to Degrees first. Then tan(60) gives about 1.732, which is $\sqrt3$.

Similar trianglesTriangles with the same angles, so one is a bigger or smaller copy of the other with every side multiplied by the same number.Read the lesson have the same angles, so matching angles have the same sine, cosine and tangent at any size. The lines, angles, and triangles lesson shows how to tell which corners match.

Say triangle $PQR$ is similar to triangle $STU$, with $P$ matching $S$ and right angles at $Q$ and $T$, and $\sin P=\frac{60}{109}$.

The question may ask about the other acute angle, $U$, which matches $R$. The side across from $P$ touches $R$, so $\cos R=\frac{60}{109}$, and so is $\cos U$. For $\tan U$, find the third side first. It's $\sqrt{109^2-60^2}=91$, so $\tan U=\frac{91}{60}$.

A ratio gives you a triangle's shape but not its size. $\cos D=\frac{5}{13}$ means the adjacent side and the hypotenuse are 5 parts and 13 parts, so call them $5k$ and $13k$. Another fact, like the perimeter, tells you $k$.

In right triangle $DEF$, the right angle is at $E$, $\cos D=\frac{5}{13}$, and the perimeter is 90. How long is $DF$?

1. $DF$ is across from the right angle. $DE$ touches $D$.

   $$DE=5k,\ DF=13k$$

2. Subtract to find the third side.

   $$EF^2=(13k)^2-(5k)^2$$

3. Square the number and the $k$ in each.

   $$EF^2=169k^2-25k^2=144k^2$$

4. Take the square root.

   $$EF=12k$$

5. Add the three sides for the perimeter.

   $$5k+12k+13k=90$$

6. Divide both sides by 30.

   $$30k=90,\ k=3$$

7. Put $k$ back in.

   $$DF=13\cdot3=39$$

The figure might label the sides with a letter instead. Say the opposite side is $t$, the adjacent side is $t+7$, and $\tan D=\frac34$.

1. Opposite over adjacent.

   $$\frac{t}{t+7}=\frac34$$

2. Cross-multiply.

   $$4t=3(t+7)$$

3. Multiply out the 3.

   $$4t=3t+21$$

4. Subtract $3t$ from both sides.

   $$t=21$$

Go back to the triangle with sides 12, 16 and 20. The side across from $L$ touches $N$, so $\sin L$ and $\cos N$ are both $\frac{12}{20}$. Angles $L$ and $N$ add to $90^\circ$, which makes them complementaryTwo angles that add up to 90 degrees.Read the lesson. The sine of an acute angle always equals the cosine of its complement.

Say acute angles $J$ and $K$ have $\sin J=\cos K$. Angle $J$ measures $(4k-6)^\circ$ and angle $K$ measures $(2k+12)^\circ$. What's the measure of the smaller angle?

1. The angles add to $90^\circ$.

   $$(4k-6)+(2k+12)=90$$

2. Add the $k$ terms and the numbers.

   $$6k+6=90$$

3. Subtract 6, then divide by 6.

   $$6k=84,\ k=14$$

4. Put 14 into the first measure.

   $$J=4(14)-6=50$$

5. And into the second. $50+40=90$, so it checks.

   $$K=2(14)+12=40$$

The smaller angle is $40^\circ$.

In Desmos, type (4x-6)+(2x+12)=90, with $x$ for $k$, and click where the vertical line crosses the axis.

If the angles are in radians, they add to $\frac{\pi}{2}$ instead, since $90^\circ$ is $\frac{\pi}{2}$ radians.

Now try one that asks for the larger angle.

+−100%Reset⤢

In triangle $QTR$, the measure of angle $Q$ is $(3x+4)^\circ$ and the measure of angle $R$ is $(x+10)^\circ$. Angles $Q$ and $R$ are acute, and $\cos Q=\sin R$.

What is the measure, in degrees, of the larger of angles $Q$ and $R$?

1. A

   $90$

2. B

   $61$

3. C

   $29$

4. D

   $13$

Show the answer

• A

  $90$

  Because $Q+R=90^\circ$, the third angle, $T$, measures $180^\circ-90^\circ=90^\circ$. The question asks for the larger of angles $Q$ and $R$, not angle $T$.

• B

  $61$

  Correct answer

  Since $Q$ and $R$ are acute, $\cos Q=\sin R$ gives $Q+R=90^\circ$. Thus $(3x+4)+(x+10)=90$, so $x=19$. The angles measure $Q=61^\circ$ and $R=29^\circ$, so the larger angle measures $61^\circ$.

• C

  $29$

  When $x=19$, angle $R$ measures $19+10=29^\circ$, which is the smaller of the two angles. The question asks for the larger angle, $Q=3(19)+4=61^\circ$.

• D

  $13$

  This comes from setting $3x+4=x+10$, which gives $x=3$ and an angle of $13^\circ$. For acute angles, $\cos Q=\sin R$ means $Q+R=90^\circ$, not $Q=R$.

Solution

The complementary-angle identity gives $\sin R=\cos(90^\circ-R)$. Since both angles are acute, $\cos Q=\sin R$ implies $Q=90^\circ-R$, so $Q+R=90^\circ$.

Substitute the angle expressionsNumbers and letters combined with no equals sign, like 5x + 7.Read the lesson: $(3x+4)+(x+10)=90$. Combine terms to get $4x+14=90$, then $4x=76$, so $x=19$.

The angle measures are $Q=3(19)+4=61^\circ$ and $R=19+10=29^\circ$. The larger angle measures $61^\circ$.

Report a problem

Some questions give an expressionNumbers and letters combined with no equals sign, like 5x + 7.Read the lesson with the sine and cosine of two complementary angles and ask which choice is equal to it. You need one more fact. For any angle $\theta$, $(\sin\theta)^2+(\cos\theta)^2=1$, which you'll also see written $\sin^2\theta+\cos^2\theta=1$.

1. Start. $35+55=90$.

   $$(\sin35^\circ)^2+(\sin55^\circ)^2$$

2. The sine of an angle is the cosine of its complement.

   $$\sin55^\circ=\cos35^\circ$$

3. Swap it in, so one angle is left.

   $$(\sin35^\circ)^2+(\cos35^\circ)^2$$

4. Sine squared plus cosine squared.

   $$=1$$

The right choice might not say 1. It can be any expression with the same value, like $(\cos55^\circ)^2+(\sin55^\circ)^2$, so rewrite each choice the same way until one matches.

On a grid-inA question where you type your own answer instead of picking one of four.Read the lesson, you can enter $\frac{7}{9}$ as 7/9, .7777 or .7778. Type 7/9 in Desmos to see the decimal. You can't enter a square root in the box.

In a story, write down the length you found with its units, then finish the rest, like a cost per mile. A count of how many things fit rounds down.

Before you pick, reread the last line. It might want the other angle, or a total built from the length you found.

Before each one, find the right angle and say what the question wants.