Geometry · G.SRT.C.8

Solving right triangles

You know one acute angle of a right triangle and the length of one side. How do you find the other sides without measuring them?

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Sine, cosine and tangentGeometry

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Angles of elevation and depressionGeometry

Where you are 7 in Right triangle trigonometry

Trigonometry7

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▴ 2 above Inverse trigonometric ratios Secant, cosecant and cotangent ratios

Solving right triangles

Trig values of the special right triangles Angles of elevation and depression

When you reach for it

A right triangle has six parts: three sides and three angles. One of the angles is already known, because it is the right angle. Tell someone one of the acute angles, which side you know (the hypotenuse, or the leg opposite or next to that angle) and its length, and the triangle is fixed: any two right triangles that share those facts are congruent. So the other two sides are not up for grabs. They have definite lengths, and this lesson is how to compute them.

What makes that possible is Similarity. Every right triangle with a 35^\circ angle has the same shape, so in every one of them the side opposite the 35^\circ angle divided by the hypotenuse is the same number. That number is \sin 35^\circ, and a calculator knows it. The ratios of sine, cosine and tangent carry the shape; the one side you know carries the size. Put them together in one equation and the unknown side drops out.

Solving a right triangle means finding every side and angle that was not given. With one acute angle and one side, the steps below find all of them. With two sides and no acute angle, the third side comes from the Pythagorean theorem and the angles come from the inverse ratios, which is a different move.

The procedure

Take the triangle in the figure: the right angle is at C, \angle A = 35^\circ, and the hypotenuse AB is 12. The legs BC = x and AC = y are wanted.

Figure 1The triangle in the procedure. The angle at A is 35^\circ and the hypotenuse is 12.

Name the sides from the known angle

Stand at the angle you know, here A, and name the three sides from there. AB, across from the right angle, is the hypotenuse. BC is opposite A. AC runs from A to the right angle, so it is adjacent to A.

Choose the ratio that holds the known side and the wanted side

Each of the three ratios uses exactly two of the three sides:

\begin{aligned} \sin A &= \frac{\text{opposite}}{\text{hypotenuse}} \\ \cos A &= \frac{\text{adjacent}}{\text{hypotenuse}} \\ \tan A &= \frac{\text{opposite}}{\text{adjacent}} \end{aligned}

and each pair of sides turns up in exactly one of them. So whichever side you know and whichever side you want, exactly one of the three ratios contains both:

Known and wanted sides	Ratio to use
opposite and hypotenuse	\sin A
adjacent and hypotenuse	\cos A
opposite and adjacent	\tan A

It does not matter which of the two is the known one. That decides where the unknown lands in the equation, not which equation to write. To find x, the side opposite A, from the hypotenuse 12, the ratio is the sine.

Write the equation and solve it

\sin 35^\circ = \frac{x}{12}

The left side is not a mystery: it is a fixed number, about 0.5736, so this is a plain linear equation in x. Multiply both sides by 12, which keeps the equation true:

x = 12 \sin 35^\circ \approx 6.88

In the same way, y is adjacent to A and the hypotenuse is known, so the cosine holds both:

\cos 35^\circ = \frac{y}{12} \quad\Longrightarrow\quad y = 12 \cos 35^\circ \approx 9.83

When the unknown lands underneath. Suppose instead the side opposite 35^\circ were known to be 6 and the hypotenuse h were wanted. The same ratio holds both, but now the unknown is in the denominator:

\sin 35^\circ = \frac{6}{h}

Multiply both sides by h (a length, so not 0) to get h \sin 35^\circ = 6, then divide both sides by \sin 35^\circ, which is not 0 either:

h = \frac{6}{\sin 35^\circ} \approx 10.46

That is the whole of the algebra. The unknown on top means multiply the ratio by the known side; the unknown underneath means divide the known side by the ratio. A quick sense check tells you which way round you have it: the hypotenuse is the longest side, so a hypotenuse that comes out shorter than a leg means the division was upside down.

Find the other angle

The three angles of a triangle add to 180^\circ (Triangle angle sum), and the right angle takes 90^\circ of that, so the two acute angles add to 90^\circ:

\angle B = 90^\circ - 35^\circ = 55^\circ

No trigonometry is needed for this part.

Keep the calculator’s digits, round only the answers

\sin 35^\circ is not exactly 0.5736; it is an endless decimal. Each answer should be computed from the calculator’s full value and rounded once, at the end. Where you can, compute every side from the given data rather than from a side you found and rounded earlier, so that one rounding does not feed the next. Example 5 shows a case where it changes the answer.

The calculator must be in degree mode. In radian mode, \sin 35 means the sine of 35 radians, which is about -0.428. A negative value for the ratio of two lengths is the giveaway.

Check

The Pythagorean theorem is an independent check: 6.88^2 + 9.83^2 = 143.96\ldots, close to 12^2 = 144, with the gap coming only from rounding. The larger acute angle, 55^\circ at B, should sit opposite the longer leg, and it does: AC \approx 9.83 is longer than BC \approx 6.88.

Worked examples

All lengths are given to two decimal places unless the example says otherwise.

Example 1

The unknown on top

Each part describes a right triangle with one acute angle and one side. Find the side asked for.

(a) The angle is 50^\circ, the leg adjacent to it is 8; find the leg opposite it.

(b) The angle is 20^\circ, the hypotenuse is 15; find the leg adjacent to it.

Solution

In each part, decide which two sides are involved, pick the ratio that holds those two, and solve. Here the wanted side always lands in the numerator, so each answer is a multiplication.

(a) Opposite and adjacent are the two sides of the tangent:

\tan 50^\circ = \frac{x}{8} \quad\Longrightarrow\quad x = 8 \tan 50^\circ \approx 9.53

The opposite leg is longer than the adjacent one, which fits: 50^\circ is more than half of a right angle, and \tan 50^\circ is more than 1.

(b) Adjacent and hypotenuse are the two sides of the cosine:

\cos 20^\circ = \frac{x}{15} \quad\Longrightarrow\quad x = 15 \cos 20^\circ \approx 14.10

A small angle at the vertex means the adjacent leg runs almost the full length of the hypotenuse, and 14.10 is just short of 15.

\sin 64^\circ = \frac{x}{5.2} \quad\Longrightarrow\quad x = 5.2 \sin 64^\circ \approx 4.67

Example 2

The unknown underneath

In a right triangle, an acute angle is 28^\circ and the leg opposite it is 9. Find the hypotenuse and the other leg.

Solution

For the hypotenuse h: the known side is opposite and the wanted side is the hypotenuse, so the sine holds both. This time the unknown is in the denominator.

\sin 28^\circ = \frac{9}{h}

Multiply both sides by h, then divide both sides by \sin 28^\circ:

h = \frac{9}{\sin 28^\circ} \approx 19.17

For the other leg a, which is adjacent to the 28^\circ angle: it could be found from the hypotenuse just computed, but that uses a rounded number. The tangent links it straight to the given leg instead:

\tan 28^\circ = \frac{9}{a} \quad\Longrightarrow\quad a = \frac{9}{\tan 28^\circ} \approx 16.93

Check. The hypotenuse should be the longest side, and 19.17 is. With Pythagoras, 9^2 + 16.93^2 = 81 + 286.6\ldots = 367.6\ldots and 19.17^2 = 367.4\ldots, which agree to rounding.

Example 3

Solving the whole triangle

In triangle ABC the right angle is at C, \angle B = 62^\circ and BC = 7. Solve the triangle.

Figure 2Example 3. Only \angle B and the side BC are known.

Solution

Three things are unknown: \angle A, AC and AB.

The angle. The acute angles add to 90^\circ, so \angle A = 90^\circ - 62^\circ = 28^\circ.

The sides. The known angle is at B, so name the sides from B. BC runs from B to the right angle, so it is adjacent to B; AC is opposite B; AB is the hypotenuse.

AC is opposite and BC is adjacent, so the tangent:

\tan 62^\circ = \frac{AC}{7} \quad\Longrightarrow\quad AC = 7 \tan 62^\circ \approx 13.17

AB is the hypotenuse and BC is adjacent, so the cosine, with the unknown underneath:

\cos 62^\circ = \frac{7}{AB} \quad\Longrightarrow\quad AB = \frac{7}{\cos 62^\circ} \approx 14.91

Both came from the given side 7, not from each other.

Check. 7^2 + 13.17^2 = 222.4\ldots and 14.91^2 = 222.3\ldots, equal to rounding. The longer leg, AC, is opposite the larger acute angle, 62^\circ.

The work could just as well have been done from \angle A = 28^\circ, naming the sides from A instead. Then BC is opposite and \displaystyle AB = \frac{7}{\sin 28^\circ}, the same number, because \sin 28^\circ = \cos 62^\circ: the side opposite one acute angle is the side adjacent to the other.

Example 4

A ladder against a wall

A ladder leans against a vertical wall on level ground. Its foot is 1.5 meters from the wall, and it makes an angle of 72^\circ with the ground. How long is the ladder, and how high up the wall does it reach?

Solution

The wall, the ground and the ladder make a right triangle, with the right angle where the wall meets the ground. The ladder is the hypotenuse. The known angle is at the foot of the ladder, so the 1.5 meters along the ground is adjacent to it, and the height up the wall is opposite it.

For the length L, adjacent and hypotenuse means the cosine:

\cos 72^\circ = \frac{1.5}{L} \quad\Longrightarrow\quad L = \frac{1.5}{\cos 72^\circ} \approx 4.85 \text{ m}

For the height h, opposite and adjacent means the tangent, again from the given 1.5:

\tan 72^\circ = \frac{h}{1.5} \quad\Longrightarrow\quad h = 1.5 \tan 72^\circ \approx 4.62 \text{ m}

The height is a little less than the ladder’s length, as it must be: the ladder is the hypotenuse.

Example 5

Two right triangles sharing a side

In triangle ABC, \angle A = 40^\circ, \angle C = 25^\circ and AB = 10. The segment BD is perpendicular to AC, with D on AC. Find BC and AC to one decimal place.

Figure 3Example 5. The segment BD splits the triangle into two right triangles that share it.

Solution

Triangle ABC has no right angle, so no ratio applies to it directly. The segment BD cuts it into two right triangles, ABD and CBD, both with the right angle at D, and they share the side BD. Work in the triangle where you know a side, then carry the shared side across.

In triangle ABD. AB = 10 is the hypotenuse and BD is opposite the 40^\circ angle:

BD = 10 \sin 40^\circ \approx 6.428

AD is adjacent to the same angle, so AD = 10 \cos 40^\circ \approx 7.660.

In triangle CBD. Now BD is the known side, opposite the 25^\circ angle at C. BC is the hypotenuse, so the sine, with the unknown underneath:

BC = \frac{BD}{\sin 25^\circ} = \frac{10 \sin 40^\circ}{\sin 25^\circ} \approx 15.2

DC is adjacent to the 25^\circ angle, so the tangent: \displaystyle DC = \frac{BD}{\tan 25^\circ} \approx 13.785. Then AC = AD + DC \approx 7.660 + 13.785 = 21.445, so AC \approx 21.4.

Why BD was not rounded. Had BD been rounded to 6.4 before the second step, the answer would have come out \displaystyle \frac{6.4}{\sin 25^\circ} \approx 15.1, not 15.2. Dividing by \sin 25^\circ \approx 0.42 more than doubles the rounding error, and here that is enough to move the last digit. Writing \displaystyle BC = \frac{10 \sin 40^\circ}{\sin 25^\circ} and evaluating once avoids it.

Example 6

Splitting an isosceles triangle

Triangle PQR has PQ = PR = 20 and \angle P = 50^\circ. Find its base QR, its height from P, and its area.

Figure 4Example 6. The altitude from P cuts the 50^\circ angle into two angles of 25^\circ.

Solution

Drop the perpendicular from P to QR, meeting it at M. The right triangles PMQ and PMR have equal hypotenuses, PQ = PR, and share the leg PM, so they are congruent (hypotenuse and leg). Their matching parts are equal: QM = MR, so M is the midpoint of QR, and \angle QPM = \angle RPM, so each is half of 50^\circ, that is 25^\circ.

Now work in triangle PMQ: the right angle is at M, the angle at P is 25^\circ, and the hypotenuse is 20. QM is opposite the 25^\circ angle and PM is adjacent to it:

QM = 20 \sin 25^\circ, \qquad PM = 20 \cos 25^\circ

So the base is QR = 2 \cdot QM = 40 \sin 25^\circ \approx 16.90, and the height is PM = 20 \cos 25^\circ \approx 18.13.

The area is half the base times the height, computed from the unrounded values:

\frac{1}{2} \cdot 40 \sin 25^\circ \cdot 20 \cos 25^\circ \approx 153.21

The base angles are equal, because the two halves are congruent, and they add to 180^\circ - 50^\circ = 130^\circ, so each is 65^\circ.

Example 7

Two unknown legs tied together

A right triangle has an acute angle of 37^\circ, and its two legs add up to 14. Find the legs.

Solution

No single side is known, so no single ratio finishes the job. Name both legs: a opposite the 37^\circ angle and b adjacent to it. The tangent ties them together, and the sum is the second fact:

\frac{a}{b} = \tan 37^\circ, \qquad a + b = 14

From the first, a = b \tan 37^\circ. Substitute into the second, and treat \tan 37^\circ as the number it is:

\begin{aligned} b + b \tan 37^\circ &= 14 \\ b\,(1 + \tan 37^\circ) &= 14 \\ b &= \frac{14}{1 + \tan 37^\circ} \approx 7.98 \end{aligned}

and a = 14 - b \approx 6.02.

Check. \displaystyle \frac{6.02}{7.98} is about 0.754, and \tan 37^\circ \approx 0.754. The hypotenuse is \displaystyle \frac{b}{\cos 37^\circ} \approx 10.00, so this triangle is very nearly the 6, 8, 10 triangle. That is no accident: the smaller acute angle of a 3, 4, 5 triangle is about 36.87^\circ, just under 37^\circ.

Boundaries

The triangle must have a right angle. The ratios are defined from a right triangle, and in any other triangle “opposite over hypotenuse” means nothing. A triangle without one can sometimes be cut into two that have one, as in Example 5; the laws of sines and cosines do that cut once and for all.

The known angle must be one of the acute angles. Knowing the right angle tells you nothing new. An acute angle lies strictly between 0^\circ and 90^\circ, and every such angle gives a real triangle.

At least one side must be known. The angles alone fix the shape and not the size: every right triangle with a 35^\circ angle is similar to every other, from tiny to huge, so no length can be found.

Two sides with no acute angle is a different move. The third side comes from the Pythagorean theorem, and the angles from the inverse ratios.

There is never a choice of triangle. One acute angle and one named side give exactly one right triangle, so every answer this method returns is the answer. The side must be named, though: “a side of 9” could be either leg or the hypotenuse, and each gives a different triangle.

Most answers are decimals, and rounded. \sin 35^\circ is irrational, so lengths found this way are approximations, unless the angle is one of the few with exact values, worked out in Trig values of the special right triangles. Round once, at the end, and say how far.

Degrees and radians are different units. The same button gives a different answer in each mode. Every angle in this lesson is in degrees.

Where this goes next

The most common use is measuring what cannot be reached: a height from a distance along the ground and an angle, as with the ladder. When one sighting is not enough, two sightings from two places pin the height down, which is the next lesson. The same triangles, laid on a map, track a ship or a plane through bearings.

Example 5 is the seed of the laws of sines and cosines. Dropping a perpendicular and solving the two right triangles works for any triangle, and doing it once with letters gives a rule that skips the perpendicular. Example 6 is the seed of the apothem and the area of a regular polygon: a regular polygon is a ring of isosceles triangles, each split the same way.

Write Exercise 18’s legs, c \sin \theta and c \cos \theta, into the Pythagorean theorem and divide by c^2, and you have the Pythagorean identity. Set c = 1 and the legs are \cos \theta and \sin \theta themselves, which is where sine and cosine on the unit circle begin.

Exercises

Give lengths to two decimal places unless the exercise says otherwise.

The unknown on top

An acute angle of a right triangle is 42^\circ and the hypotenuse is 20. Find the leg opposite the angle.
An acute angle is 58^\circ and the hypotenuse is 9. Find the leg adjacent to the angle.
An acute angle is 15^\circ and the leg adjacent to it is 30. Find the leg opposite it.

The unknown underneath

An acute angle is 33^\circ and the leg opposite it is 11. Find the hypotenuse.
An acute angle is 47^\circ and the leg adjacent to it is 6. Find the hypotenuse.
An acute angle is 71^\circ and the leg opposite it is 25. Find the leg adjacent to it.

Solve the triangle

In each, the right angle of triangle ABC is at C. Find the missing angle and both missing sides.

\angle A = 24^\circ and AB = 16.
\angle B = 53^\circ and AC = 12.
\angle A = 68^\circ and BC = 4.5.

More than one triangle

An isosceles triangle has two sides of 14 and base angles of 70^\circ. Find its base and its height to the base.
The diagonal of a rectangle is 25 and makes an angle of 31^\circ with the longer side. Find the two sides and the area.
In triangle ABC, \angle A = 55^\circ, \angle C = 30^\circ and AB = 8. The segment BD is perpendicular to AC, with D on AC. Find BD, BC and AC to one decimal place.
A regular pentagon has sides of 10. Find its apothem (the distance from the center to the middle of a side) and its radius (the distance from the center to a corner).

Problems

A ramp rises 0.75 m and makes an angle of 4.8^\circ with the level ground. How long is the ramp’s surface, and how far does it run along the ground?
A kite string is 60 m long, held taut and straight, and makes an angle of 52^\circ with the horizontal. The hand holding it is 1.2 m above the ground. How high is the kite, to one decimal place?
A pole on level ground casts a shadow 9.4 m long when the sun’s rays meet the ground at 38^\circ. How tall is the pole?
A plane climbs in a straight line at 12^\circ to the level ground. After it has flown 5 km along that line, how high is it, and how far has it moved over the ground?

Using letters

A right triangle has an acute angle \theta and hypotenuse c. Write the leg opposite \theta and the leg adjacent to \theta in terms of c and \theta.
A right triangle has an acute angle of 40^\circ, and its longer leg is 3 more than its shorter leg. Find both legs.

Reasoning

To find the hypotenuse from an angle and the leg opposite it, you divide the leg by the sine instead of multiplying. Explain why, and explain how the length of the answer tells you if it was done the wrong way round.
Someone finds one leg of a right triangle, rounds it, then uses the rounded leg to find the hypotenuse. Explain why their hypotenuse can differ in its last digit from one computed straight from the given data.
Explain why the three angles of a right triangle, with no side, are not enough to find any of its sides.

Answers

13.38. Opposite and hypotenuse: \displaystyle \sin 42^\circ = \frac{x}{20}, so x = 20 \sin 42^\circ \approx 13.38.
4.77. Adjacent and hypotenuse: \displaystyle \cos 58^\circ = \frac{x}{9}, so x = 9 \cos 58^\circ.
8.04. Opposite and adjacent: \displaystyle \tan 15^\circ = \frac{x}{30}, so x = 30 \tan 15^\circ.
20.20. \displaystyle \sin 33^\circ = \frac{11}{h}, so \displaystyle h = \frac{11}{\sin 33^\circ}.
8.80. \displaystyle \cos 47^\circ = \frac{6}{h}, so \displaystyle h = \frac{6}{\cos 47^\circ}.
8.61. \displaystyle \tan 71^\circ = \frac{25}{a}, so \displaystyle a = \frac{25}{\tan 71^\circ}. It is the shorter leg, opposite the smaller angle, 19^\circ.
\angle B = 66^\circ, BC \approx 6.51, AC \approx 14.62. BC = 16 \sin 24^\circ is opposite A, and AC = 16 \cos 24^\circ is adjacent to it.
\angle A = 37^\circ, BC \approx 9.04, AB \approx 15.03. AC is opposite B and BC is adjacent to it, so \displaystyle BC = \frac{12}{\tan 53^\circ}, and \displaystyle AB = \frac{12}{\sin 53^\circ}.
\angle B = 22^\circ, AC \approx 1.82, AB \approx 4.85. BC is opposite A, so \displaystyle AC = \frac{4.5}{\tan 68^\circ} and \displaystyle AB = \frac{4.5}{\sin 68^\circ}.
Base \approx 9.58, height \approx 13.16. The height to the base splits the triangle into two right triangles with hypotenuse 14 and a 70^\circ angle at the base. The height is opposite that angle, 14 \sin 70^\circ; half the base is adjacent, 14 \cos 70^\circ, so the base is 28 \cos 70^\circ.
Sides \approx 21.43 and \approx 12.88, area \approx 275.92. The diagonal is the hypotenuse of a right triangle whose legs are the sides: 25 \cos 31^\circ along the longer side and 25 \sin 31^\circ across. The area is their product, from the unrounded values.
BD \approx 6.6, BC \approx 13.1, AC \approx 15.9. BD = 8 \sin 55^\circ \approx 6.553. \displaystyle BC = \frac{BD}{\sin 30^\circ}. AD = 8 \cos 55^\circ \approx 4.589 and \displaystyle DC = \frac{BD}{\tan 30^\circ} \approx 11.351, so AC = AD + DC.
Apothem \approx 6.88, radius \approx 8.51. Join the center to the two ends of one side and to its midpoint. The central angle is \displaystyle \frac{360^\circ}{5} = 72^\circ, and the two radii are equal, so, as in Example 6, the apothem halves it, making a right triangle with a 36^\circ angle at the center and the half side 5 opposite it. Apothem \displaystyle = \frac{5}{\tan 36^\circ}, radius \displaystyle = \frac{5}{\sin 36^\circ}.
Surface \approx 8.96 m, along the ground \approx 8.93 m. The rise is opposite the angle: surface \displaystyle = \frac{0.75}{\sin 4.8^\circ}, ground \displaystyle = \frac{0.75}{\tan 4.8^\circ}. At so small an angle the two are nearly equal.
48.5 m. The string is the hypotenuse and the rise above the hand is opposite the angle: 60 \sin 52^\circ \approx 47.28. Add the 1.2 m of the hand.
7.34 m. The shadow is adjacent to the 38^\circ angle and the pole is opposite it: 9.4 \tan 38^\circ.
Height \approx 1.04 km, over the ground \approx 4.89 km. The 5 km is the hypotenuse: 5 \sin 12^\circ and 5 \cos 12^\circ.
Opposite = c \sin \theta, adjacent = c \cos \theta. Multiply both sides of \displaystyle \sin \theta = \frac{\text{opposite}}{c} and of \displaystyle \cos \theta = \frac{\text{adjacent}}{c} by c.
18.65 and 15.65. Since \tan 40^\circ < 1, the leg opposite 40^\circ, call it a, is the shorter one, and the adjacent leg is b = a + 3. Then a = b \tan 40^\circ, so b - b \tan 40^\circ = 3 and \displaystyle b = \frac{3}{1 - \tan 40^\circ} \approx 18.65, with a = b - 3 \approx 15.65.
The equation is \displaystyle \sin A = \frac{\text{opposite}}{h} with h in the denominator. Multiplying both sides by h gives h \sin A = \text{opposite}, and dividing by \sin A gives \displaystyle h = \frac{\text{opposite}}{\sin A}. Since 0 < \sin A < 1 for an acute angle, dividing by it makes the answer longer than the leg, as a hypotenuse must be; multiplying would make it shorter, which is impossible for a hypotenuse.
The rounded leg is off from the true leg by up to half a unit in its last place. The next step multiplies or divides that error along with the leg, and dividing by a ratio less than 1 makes it bigger. If the enlarged error crosses a rounding boundary, the last digit changes. Computing from the given data rounds only once.
Any two right triangles with the same angles are similar: one is a scaled copy of the other. Scaling by any factor keeps every angle and multiplies every side, so infinitely many triangles, of every size, share those angles. The ratios only fix how the sides compare to each other, and a single known length is what fixes the scale.