Teach Me Maths

Pythagoras and Trigonometry in 2D

Choose Pythagoras or a trigonometric ratio in right-angled triangles, use exact values and Pythagorean triples, and combine triangles in 2D problems.

Identify the right-angled triangle first

For a right-angled triangle with hypotenuse c:

a^2+b^2=c^2

For an acute angle \theta:

\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}

\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}

\tan\theta=\frac{\text{opposite}}{\text{adjacent}}

Label opposite and adjacent relative to the angle you are using. The hypotenuse is always opposite the right angle. Use Pythagoras when two sides are known; use a trig ratio when the information involves an acute angle. Work in degrees.

Explore a right-angled triangle

As the angle increases, its opposite side grows and its adjacent side shrinks. The labels “opposite” and “adjacent” swap if you use the other acute angle.

Worked example 1

A right-angled triangle has hypotenuse 25 cm and one shorter side 7 cm. Find its other side.

  1. Let the missing side be b. Then 7^2+b^2=25^2, so b^2=25^2-7^2

  2. Your turn. Calculate 25^2-7^2

    b^2=576

  3. Answer: b=24 cm. The positive root gives a side shorter than the hypotenuse.

Worked example 2

For positive integers m>n, show that m^2-n^2, 2mn and m^2+n^2 form a Pythagorean triple. Then find the triple when m=4 and n=1.

  1. All three lengths are positive integers, and m^2+n^2 is the largest. Expand the sum of the squares of the shorter two.

  2. Your turn. In (m^2-n^2)^2=m^4+kn^2m^2+n^4, what is k?

    (m^2-n^2)^2=m^4-2m^2n^2+n^4

  3. Adding (2mn)^2=4m^2n^2 gives m^4+2m^2n^2+n^4=(m^2+n^2)^2. This proves Pythagoras holds.
  4. Answer: the expressions always form a Pythagorean triple; for m=4,n=1 it is (15,8,17). Multiplying all three lengths in any triple by the same positive integer gives another triple.

Worked example 3

A ladder of length 6.5 m rests against a vertical wall on horizontal ground. It makes an angle of 68^\circ with the ground. Find the height reached, to 3 significant figures.

  2. \sin68^\circ=h/6.5, so h=6.5\sin68^\circ

  3. Your turn. Evaluate the height in metres to 3 significant figures.

    h=6.026695\ldots m.

  4. Answer: 6.03 m (3 s.f.). The vertical height is less than the ladder’s length.

Worked example 4

In a right-angled triangle, the side opposite an acute angle \theta is 5 cm and the hypotenuse is 13 cm. Find \theta to 1 decimal place.

  1. \sin\theta=5/13, so \theta=\sin^{-1}(5/13). Here \sin^{-1} means inverse sine, not a reciprocal.

  2. Your turn. Give the angle in degrees to 1 decimal place.

    \theta=22.619864\ldots^\circ

  3. Answer: 22.6^\circ (1 d.p.). Only the acute value fits a right-angled triangle’s remaining angle.

Exact values from special triangles

Bisect an equilateral triangle of side 2. Its half has sides 1,\sqrt3,2 opposite angles 30^\circ,60^\circ,90^\circ. Thus:

\sin30^\circ=\cos60^\circ=\frac12

\cos30^\circ=\sin60^\circ=\frac{\sqrt3}{2}

\tan30^\circ=\frac1{\sqrt3}

\tan60^\circ=\sqrt3

A right isosceles triangle with legs 1,1 has hypotenuse \sqrt2. Therefore:

\sin45^\circ=\cos45^\circ=\frac1{\sqrt2}=\frac{\sqrt2}{2}

\tan45^\circ=1

Worked example 5

A right-angled triangle has hypotenuse 12 cm and an angle of 30^\circ. Find both shorter sides exactly.

  1. Your turn. The side opposite 30^\circ is 12\sin30^\circ. Find its length in cm.

    12\times\frac12=6 cm.

  2. The adjacent side is 12\cos30^\circ=12\times\frac{\sqrt3}{2}=6\sqrt3 cm.
  3. Answer: 6 cm opposite 30^\circ and 6\sqrt3 cm adjacent. These follow the ratio 1:\sqrt3:2

Worked example 6

A walker goes 8 km due east, then 10 km on a bearing of 030^\circ. Find the straight-line distance from the start exactly and the bearing from the start to the finish to 1 decimal place.

  2. The second leg adds 10\sin30^\circ=5 km east and 10\cos30^\circ=5\sqrt3 km north. Total displacement is 13 km east and 5\sqrt3 km north.

  3. Your turn. Calculate the squared distance: 13^2+(5\sqrt3)^2

    The distance is \sqrt{244}=2\sqrt{61} km.

  4. For the angle clockwise from north, \tan\beta=13/(5\sqrt3), so \beta=56.3295\ldots^\circ
  5. Answer: distance 2\sqrt{61} km; bearing 056.3^\circ (1 d.p.). Measuring from east instead would give the wrong bearing.

Worked example 7

An isosceles triangle has equal sides 13 cm and base 10 cm. Find its area and its apex angle to 1 decimal place.

  2. Your turn. Find the perpendicular height \sqrt{13^2-5^2} in cm.

    The height is 12 cm.

  3. Area =\frac12(10)(12)=60 cm². Half the apex angle is \sin^{-1}(5/13)=22.6198\ldots^\circ
  4. Answer: area 60 cm² and apex angle 2\sin^{-1}(5/13)=45.2^\circ (1 d.p.). Double the unrounded half-angle.

Common mistakes

  • Using Pythagoras or SOHCAHTOA without a right angle. Check that the triangle is right-angled first.
  • Changing the hypotenuse. It stays opposite the right angle; only opposite and adjacent swap.
  • Using the whole base in a split isosceles triangle. Use half the base in each smaller triangle; the whole base for the original area.
  • Measuring a bearing from east. Measure clockwise from north at the starting point.
  • Rounding intermediate values. Keep full precision until the final answer, including when doubling a half-angle.

Now try it: Pythagoras and Trigonometry in 2D practice questions

More on this topic: Pythagoras and Trigonometry in 2D worksheet with full solutions