Right triangle trigonometry

Print the chapter

Every lesson of the chapter, ready for paper. Exercise sets are lettered A, B, C… and answered at the end.

To save a PDF, choose “Save as PDF” in the print dialog.

1. Special right triangles

Warm-up

Three quick questions from earlier lessons.

Discover

A square with side 1 is cut along its diagonal. How long is the diagonal?

  1. 2\sqrt{2}
  2. 22

Key idea

Two triangles whose sides you can know exactly.

45-45-90 (half a square): sides 1:1:21 : 1 : \sqrt{2}. 30-60-90 (half an equilateral triangle): sides 1:3:21 : \sqrt{3} : 2, with the shortest side opposite 30∘30^\circ and the hypotenuse twice as long.

Worked example

A 30-60-90 triangle with hypotenuse 10.

Short leg 55
Half the hypotenuse.
Long leg 535\sqrt{3}
Short leg × √3.

Your turn, guided

Your turn: a 45-45-90 triangle with hypotenuse 626\sqrt{2}. Each leg:

66
Divide by √2.

Practice ★

  1. An isosceles right triangle has legs of length 8. Find the hypotenuse (exact).
  2. An isosceles right triangle has legs of length 6. Find the hypotenuse (exact).
  1. In a 30-60-90 triangle the hypotenuse is 6. Find the side opposite the 30∘30^\circ angle.

Watch out!

Practice ★★

  1. In a 30-60-90 triangle the shortest side is 10. Find the longer leg (exact).
  2. In a 30-60-90 triangle the shortest side is 2. Find the longer leg (exact).

Exit ticket

Three questions to check you have got it.

  1. An isosceles right triangle has legs of length 8. Find the hypotenuse (exact).
  2. An isosceles right triangle has legs of length 6. Find the hypotenuse (exact).
  3. An isosceles right triangle has legs of length 11. Find the hypotenuse (exact).

Summary card

Special right triangles

45-45-90: 1:1:21 : 1 : \sqrt{2}. 30-60-90: 1:3:21 : \sqrt{3} : 2.

Watch out: √2 vs √3.

2. Trigonometric ratios in right triangles

Warm-up

Three quick questions from earlier lessons.

Discover

Two right triangles of different sizes both have a 35∘35^\circ angle. Is the ratio opposite/hypotenuse the same in both?

  1. Yes
  2. No, the bigger triangle has a bigger ratio

Make the triangle bigger and smaller. Divide the opposite side by the hypotenuse.

  1. It stays 0.6
  2. It grows with the triangle

Key idea

Ratios that depend only on the angle.

For an acute angle AA of a right triangle: sin⁡A=oppositehypotenuse\sin A = \dfrac{\text{opposite}}{\text{hypotenuse}}, cos⁡A=adjacenthypotenuse\cos A = \dfrac{\text{adjacent}}{\text{hypotenuse}}, tan⁡A=oppositeadjacent\tan A = \dfrac{\text{opposite}}{\text{adjacent}} (SOH-CAH-TOA).

Worked example

Legs 8 (opposite AA) and 15, hypotenuse 17.

sin⁡A=817\sin A = \tfrac{8}{17}
SOH.
cos⁡A=1517\cos A = \tfrac{15}{17}
CAH.
tan⁡A=815\tan A = \tfrac{8}{15}
TOA.

Your turn, guided

Your turn: legs 7 (opposite) and 24, hypotenuse 25. cos⁡A\cos A:

2425\tfrac{24}{25}
Adjacent/hypotenuse.

Practice ★

  1. A right triangle has legs 12 and 16 and hypotenuse 20. Angle AA is opposite the leg 12. Find sin⁡A\sin A.
  2. A right triangle has legs 6 and 8 and hypotenuse 10. Angle AA is opposite the leg 6. Find sin⁡A\sin A.
  1. In a right triangle with acute angle 21∘21^\circ, which ratio is cos⁡21∘\cos 21^\circ?
    1. adjacent ÷ hypotenuse
    2. opposite ÷ hypotenuse
    3. opposite ÷ adjacent

Watch out!

Practice ★★

  1. Same kind of triangle: legs 15 and 36, hypotenuse 39. Angle BB is opposite the leg 36. Find tan⁡B\tan B.
  2. Same kind of triangle: legs 20 and 48, hypotenuse 52. Angle BB is opposite the leg 48. Find tan⁡B\tan B.

Exit ticket

Three questions to check you have got it.

  1. A right triangle has legs 12 and 16 and hypotenuse 20. Angle AA is opposite the leg 12. Find sin⁡A\sin A.
  2. A right triangle has legs 6 and 8 and hypotenuse 10. Angle AA is opposite the leg 6. Find sin⁡A\sin A.
  3. A right triangle has legs 27 and 36 and hypotenuse 45. Angle AA is opposite the leg 27. Find sin⁡A\sin A.

Summary card

Trigonometric ratios

SOH-CAH-TOA, for an acute angle of a right triangle.

Watch out: opposite and adjacent depend on which angle you look from.

3. Finding sides with trigonometry

Warm-up

Three quick questions from earlier lessons.

Discover

A kite string 30 m long makes a 40∘40^\circ angle with the ground. Which ratio gives the kite's height?

  1. sin⁡40∘=height30\sin 40^\circ = \dfrac{\text{height}}{30}
  2. cos⁡40∘=height30\cos 40^\circ = \dfrac{\text{height}}{30}

Key idea

Pick the ratio that links what you know with what you want.

Label the sides from the given angle, choose sin, cos or tan, write the equation, then solve: if the unknown is on top, multiply; if it is in the denominator, divide. Use degree mode.

Worked example

Kite: height =30sin⁡40∘= 30 \sin 40^\circ.

30×0.64330 \times 0.643
sin 40° ≈ 0.643.
height ≈19.28\approx 19.28 m
Round.

Your turn, guided

Your turn: hypotenuse 10, angle 30∘30^\circ. The side opposite.

55
10 × sin 30° = 10 × ½.

Practice ★

  1. A right triangle has hypotenuse 14 and an acute angle 43∘43^\circ. Find the side opposite that angle (to 2 decimal places).
  2. A right triangle has hypotenuse 19 and an acute angle 22∘22^\circ. Find the side opposite that angle (to 2 decimal places).
  1. A ramp rises at 55∘55^\circ over a horizontal distance of 15 m. How high does it rise? (2 decimal places)

Watch out!

Practice ★★

  1. The side opposite a 40∘40^\circ angle in a right triangle is 4. Find the hypotenuse (2 decimal places).
  2. The side opposite a 40∘40^\circ angle in a right triangle is 9. Find the hypotenuse (2 decimal places).
  1. Boss problem. A 8 m ladder leans against a wall at 55∘55^\circ to the ground. Give answers to 2 decimal places.

Exit ticket

Three questions to check you have got it.

  1. A right triangle has hypotenuse 14 and an acute angle 43∘43^\circ. Find the side opposite that angle (to 2 decimal places).
  2. A right triangle has hypotenuse 19 and an acute angle 22∘22^\circ. Find the side opposite that angle (to 2 decimal places).
  3. A right triangle has hypotenuse 18 and an acute angle 58∘58^\circ. Find the side opposite that angle (to 2 decimal places).

Summary card

Finding sides

Label → choose the ratio → solve (multiply or divide).

Watch out: the hypotenuse is the longest side; check your answer.

4. Finding angles with inverse trigonometry

Warm-up

Three quick questions from earlier lessons.

Discover

In a right triangle, tan⁡A=1\tan A = 1. What is AA?

  1. 45∘45^\circ
  2. 1∘1^\circ

Key idea

Inverse trigonometric functions turn ratios back into angles.

If sin⁡A=r\sin A = r then A=sin⁡−1rA = \sin^{-1} r; similarly cos⁡−1\cos^{-1} and tan⁡−1\tan^{-1}. Here −1^{-1} means inverse function, not reciprocal. Use degree mode.

Worked example

Legs 5 (opposite) and 8 (adjacent).

tan⁡A=58=0.625\tan A = \tfrac{5}{8} = 0.625
TOA.
A=tan⁡−1(0.625)≈32.0∘A = \tan^{-1}(0.625) \approx 32.0^\circ
Inverse tangent.

Your turn, guided

Your turn: sin⁡A=12\sin A = \tfrac{1}{2}. Find AA.

30∘30^\circ
A special angle.

Practice ★

  1. A right triangle has legs 10 (opposite angle AA) and 12 (adjacent). Find angle AA in degrees (1 decimal place).
  2. A right triangle has legs 8 (opposite angle AA) and 9 (adjacent). Find angle AA in degrees (1 decimal place).
  1. What does sin⁡−1(0.7)\sin^{-1}(0.7) mean?
    1. The angle whose sine is 0.7
    2. 1sin⁡0.7\dfrac{1}{\sin 0.7}
    3. −sin⁡(0.7)-\sin(0.7)

Watch out!

Practice ★★

  1. In a right triangle the side opposite angle AA is 5 and the hypotenuse is 15. Find AA in degrees (1 decimal place).
  2. In a right triangle the side opposite angle AA is 6 and the hypotenuse is 10. Find AA in degrees (1 decimal place).

Exit ticket

Three questions to check you have got it.

  1. A right triangle has legs 10 (opposite angle AA) and 12 (adjacent). Find angle AA in degrees (1 decimal place).
  2. A right triangle has legs 8 (opposite angle AA) and 9 (adjacent). Find angle AA in degrees (1 decimal place).
  3. A right triangle has legs 12 (opposite angle AA) and 9 (adjacent). Find angle AA in degrees (1 decimal place).

Summary card

Finding angles

A=sin⁡−1A = \sin^{-1}, cos⁡−1\cos^{-1} or tan⁡−1\tan^{-1} of the side ratio.

Watch out: inverse, not reciprocal; degree mode.

5. Sine and cosine of complementary angles

Warm-up

Three quick questions from earlier lessons.

Discover

In a right triangle with acute angles AA and BB, the side opposite AA is adjacent to BB. So sin⁡A\sin A equals…

  1. cos⁡B\cos B
  2. sin⁡B\sin B

Key idea

Co-functions of complementary angles.

The acute angles of a right triangle add to 90∘90^\circ, and sin⁡x=cos⁡(90∘−x)\sin x = \cos(90^\circ - x), cos⁡x=sin⁡(90∘−x)\cos x = \sin(90^\circ - x).

Worked example

sin⁡25∘=cos⁡?\sin 25^\circ = \cos ?

90−25=6590 - 25 = 65
Complement.
sin⁡25∘=cos⁡65∘\sin 25^\circ = \cos 65^\circ
Co-function.

Your turn, guided

Your turn: cos⁡72∘=sin⁡y∘\cos 72^\circ = \sin y^\circ with yy acute.

y=18y = 18
90 − 72.

Practice ★

  1. Find the angle yy between 0∘0^\circ and 90∘90^\circ with sin⁡21∘=cos⁡y∘\sin 21^\circ = \cos y^\circ (degrees).
  2. Find the angle yy between 0∘0^\circ and 90∘90^\circ with sin⁡69∘=cos⁡y∘\sin 69^\circ = \cos y^\circ (degrees).
  1. In a right triangle, sin⁡A=310\sin A = \dfrac{ 3 }{ 10 }. Find cos⁡B\cos B, where BB is the other acute angle.

Watch out!

Practice ★★

  1. Why is sin⁡65∘=cos⁡25∘\sin 65^\circ = \cos 25^\circ?
    1. In one right triangle, the side opposite the 65∘65^\circ angle is adjacent to the other acute angle
    2. Because sin and cos always add to 1
    3. Because sin and cos are the same function

Exit ticket

Three questions to check you have got it.

  1. Find the angle yy between 0∘0^\circ and 90∘90^\circ with sin⁡21∘=cos⁡y∘\sin 21^\circ = \cos y^\circ (degrees).
  2. Find the angle yy between 0∘0^\circ and 90∘90^\circ with sin⁡69∘=cos⁡y∘\sin 69^\circ = \cos y^\circ (degrees).
  3. Find the angle yy between 0∘0^\circ and 90∘90^\circ with sin⁡80∘=cos⁡y∘\sin 80^\circ = \cos y^\circ (degrees).

Summary card

Complementary angles

sin⁡x=cos⁡(90∘−x)\sin x = \cos(90^\circ - x).

Watch out: equal values need complementary angles, not the same angle.

6. The Pythagorean identity and simplifying

Warm-up

Three quick questions from earlier lessons.

Discover

The point (cos⁡x,sin⁡x)(\cos x, \sin x) is on the unit circle, whose equation is X2+Y2=1X^2 + Y^2 = 1. What does that tell you?

  1. cos⁡2x+sin⁡2x=1\cos^2 x + \sin^2 x = 1
  2. cos⁡x+sin⁡x=1\cos x + \sin x = 1

Key idea

An identity is an equation that is true for every value of the variable.

sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1, so cos⁡2x=1−sin⁡2x\cos^2 x = 1 - \sin^2 x and sin⁡2x=1−cos⁡2x\sin^2 x = 1 - \cos^2 x. Also tan⁡x=sin⁡xcos⁡x\tan x = \dfrac{\sin x}{\cos x}. Use them to find one value from another (the quadrant gives the sign) and to simplify.

Worked example

cos⁡θ=−35\cos\theta = -\dfrac{3}{5} and θ\theta is in quadrant III. Find sin⁡θ\sin\theta.

sin⁡2θ=1−925=1625\sin^2\theta = 1 - \dfrac{9}{25} = \dfrac{16}{25}
The Pythagorean identity.
sin⁡θ=±45\sin\theta = \pm\dfrac{4}{5}
Two square roots are possible.
sin⁡θ=−45\sin\theta = -\dfrac{4}{5}
Sine is negative in quadrant III.

Your turn, guided

Your turn: sin⁡θ=513\sin\theta = \dfrac{5}{13} and θ\theta is in quadrant I.

cos⁡2θ=144169\cos^2\theta = \dfrac{144}{169}
1 − 25/169.
cos⁡θ=1213\cos\theta = \dfrac{12}{13}
Positive in quadrant I.

Practice ★

  1. sin⁡θ=15\sin\theta = \frac{1}{5} and θ\theta is in quadrant II. Find the exact value of cos⁡θ\cos\theta.
  2. sin⁡θ=38\sin\theta = \frac{3}{8} and θ\theta is in quadrant II. Find the exact value of cos⁡θ\cos\theta.
  3. sin⁡θ=16\sin\theta = \frac{1}{6} and θ\theta is in quadrant II. Find the exact value of cos⁡θ\cos\theta.
  4. sin⁡θ=13\sin\theta = \frac{1}{3} and θ\theta is in quadrant II. Find the exact value of cos⁡θ\cos\theta.

Watch out!

Practice ★★

  1. Which expression is equal to 1−cos⁡2xsin⁡x\dfrac{1 - \cos^2 x}{\sin x} (where sin⁡x≠0\sin x \ne 0)?
    1. sin⁡x\sin x
    2. cos⁡x\cos x
    3. 1−cos⁡x1 - \cos x
  1. Which equation is true for every xx?
    1. sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1
    2. sin⁡x+cos⁡x=1\sin x + \cos x = 1
    3. sin⁡2x−cos⁡2x=1\sin^2 x - \cos^2 x = 1
    4. tan⁡x=cos⁡xsin⁡x\tan x = \dfrac{\cos x}{\sin x}

Exit ticket

Three questions to check you have got it.

  1. sin⁡θ=15\sin\theta = \frac{1}{5} and θ\theta is in quadrant II. Find the exact value of cos⁡θ\cos\theta.
  2. sin⁡θ=38\sin\theta = \frac{3}{8} and θ\theta is in quadrant II. Find the exact value of cos⁡θ\cos\theta.
  3. sin⁡θ=16\sin\theta = \frac{1}{6} and θ\theta is in quadrant II. Find the exact value of cos⁡θ\cos\theta.

Summary card

The Pythagorean identity

  • sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1 for every xx.
  • From one value: square, subtract from 1, take the root, choose the sign by quadrant.

Watch out: the identity needs the squares.

Answers: Right triangle trigonometry

1. Special right triangles

A

  1. 828 \sqrt{2}
  2. 626 \sqrt{2}

B

  1. 33

C

  1. 10310 \sqrt{3}
  2. 232 \sqrt{3}

D

  1. 828 \sqrt{2}
  2. 626 \sqrt{2}
  3. 11211 \sqrt{2}

2. Trigonometric ratios in right triangles

A

  1. 35\frac{3}{5}
  2. 35\frac{3}{5}

B

  1. adjacent ÷ hypotenuse

C

  1. 125\frac{12}{5}
  2. 125\frac{12}{5}

D

  1. 35\frac{3}{5}
  2. 35\frac{3}{5}
  3. 35\frac{3}{5}

3. Finding sides with trigonometry

A

  1. 14sin⁡(43π180)14 \sin{\left(\frac{43 \pi}{180} \right)}
  2. 19sin⁡(11π90)19 \sin{\left(\frac{11 \pi}{90} \right)}

B

  1. 15tan⁡(11π36)15 \tan{\left(\frac{11 \pi}{36} \right)}

C

  1. 4sin⁡(2π9)\frac{4}{\sin{\left(\frac{2 \pi}{9} \right)}}
  2. 9sin⁡(2π9)\frac{9}{\sin{\left(\frac{2 \pi}{9} \right)}}

D

  1. 3535

E

  1. 14sin⁡(43π180)14 \sin{\left(\frac{43 \pi}{180} \right)}
  2. 19sin⁡(11π90)19 \sin{\left(\frac{11 \pi}{90} \right)}
  3. 18sin⁡(29π90)18 \sin{\left(\frac{29 \pi}{90} \right)}

4. Finding angles with inverse trigonometry

A

  1. 180tan⁡−1(56)π\frac{180 \tan^{-1}{\left(\frac{5}{6} \right)}}{\pi}
  2. 180tan⁡−1(89)π\frac{180 \tan^{-1}{\left(\frac{8}{9} \right)}}{\pi}

B

  1. The angle whose sine is 0.7

C

  1. 180sin⁡−1(13)π\frac{180 \sin^{-1}{\left(\frac{1}{3} \right)}}{\pi}
  2. 180sin⁡−1(35)π\frac{180 \sin^{-1}{\left(\frac{3}{5} \right)}}{\pi}

D

  1. 180tan⁡−1(56)π\frac{180 \tan^{-1}{\left(\frac{5}{6} \right)}}{\pi}
  2. 180tan⁡−1(89)π\frac{180 \tan^{-1}{\left(\frac{8}{9} \right)}}{\pi}
  3. 180tan⁡−1(43)π\frac{180 \tan^{-1}{\left(\frac{4}{3} \right)}}{\pi}

5. Sine and cosine of complementary angles

A

  1. 6969
  2. 2121

B

  1. 310\frac{3}{10}

C

  1. In one right triangle, the side opposite the 65∘65^\circ angle is adjacent to the other acute angle

D

  1. 6969
  2. 2121
  3. 1010

6. The Pythagorean identity and simplifying

A

  1. −265- \frac{2 \sqrt{6}}{5}
  2. −558- \frac{\sqrt{55}}{8}
  3. −356- \frac{\sqrt{35}}{6}
  4. −223- \frac{2 \sqrt{2}}{3}

B

  1. sin⁡x\sin x

C

  1. sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1

D

  1. −265- \frac{2 \sqrt{6}}{5}
  2. −558- \frac{\sqrt{55}}{8}
  3. −356- \frac{\sqrt{35}}{6}