Pythagoras and Trigonometry in 2D

Use the triangle with sides 1, \sqrt3 and 2: the side opposite the 30^\circ angle is the shortest side.

1. In the 1:\sqrt3:2 triangle, the side opposite 30^\circ is 1 and the side adjacent to it is \sqrt3
2. \tan 30^\circ=\dfrac{\text{opposite}}{\text{adjacent}}=\dfrac{1}{\sqrt3}
3. Rationalise: \dfrac{1}{\sqrt3}=\dfrac{\sqrt3}{3}
4. Answer: \dfrac{\sqrt3}{3} (or \dfrac{1}{\sqrt3})

Check whether the squares of the two shorter sides add up to the square of the longest side, or look for a multiple of a Pythagorean triple.

1. The sides of a right-angled triangle satisfy a^2+b^2=c^2, where c is the longest side
2. 14,\ 48,\ 50 is 2\times(7,\ 24,\ 25): 14^2+48^2=196+2304=2500=50^2
3. 9^2+40^2=1681=41^2, not 42^2; \ 10^2+24^2=676=26^2, not 25^2; \ 8^2+15^2=289=17^2, not 18^2
4. Answer: 14,\ 48,\ 50

Write down \sin 30^\circ and \cos 45^\circ from the 1:\sqrt3:2 and 1:1:\sqrt2 triangles, then divide.

1. \sin 30^\circ=\dfrac12
2. \cos 45^\circ=\dfrac{1}{\sqrt2}=\dfrac{\sqrt2}{2}
3. \dfrac12\div\dfrac{1}{\sqrt2}=\dfrac12\times\sqrt2=\dfrac{\sqrt2}{2}
4. Answer: \dfrac{\sqrt2}{2}

\sin^2 60^\circ means (\sin 60^\circ)^2: write down the exact values first, then square them.

1. \sin 60^\circ=\dfrac{\sqrt3}{2}, so \sin^2 60^\circ=\dfrac34
2. \tan 45^\circ=1, so \tan^2 45^\circ=1
3. 8\times\dfrac34-1=6-1
4. Answer: 5

BC is opposite the 30^\circ angle and AB is adjacent to it, so use \tan 30^\circ.

1. BC is opposite angle A and AB is adjacent, so \tan 30^\circ=\dfrac{BC}{12}
2. BC=12\tan 30^\circ=12\times\dfrac{1}{\sqrt3}=\dfrac{12}{\sqrt3}
3. Rationalise: \dfrac{12}{\sqrt3}\times\dfrac{\sqrt3}{\sqrt3}=\dfrac{12\sqrt3}{3}
4. Answer: 4\sqrt3 cm

PR is the hypotenuse and PQ is adjacent to angle QPR: simplify \dfrac{\sqrt6}{2\sqrt3} and compare it with the exact values you know.

1. \cos QPR=\dfrac{\text{adjacent}}{\text{hypotenuse}}=\dfrac{\sqrt6}{2\sqrt3}
2. \dfrac{\sqrt6}{\sqrt3}=\sqrt2, so \cos QPR=\dfrac{\sqrt2}{2}
3. \cos 45^\circ=\dfrac{\sqrt2}{2}
4. Answer: 45^\circ

The height splits the triangle into two right-angled triangles with a 60^\circ angle; use \sin 60^\circ=\dfrac{\sqrt3}{2}.

1. The height cuts the triangle into two right-angled triangles with angles 30^\circ, 60^\circ, 90^\circ
2. The side s is the hypotenuse and the height is opposite the 60^\circ angle
3. \sin 60^\circ=\dfrac{9}{s}, so \dfrac{\sqrt3}{2}=\dfrac{9}{s}
4. s=\dfrac{18}{\sqrt3}
5. Rationalise: \dfrac{18}{\sqrt3}\times\dfrac{\sqrt3}{\sqrt3}=\dfrac{18\sqrt3}{3}=6\sqrt3
6. Answer: 6\sqrt3 cm

The diagonals of a rhombus bisect each other at right angles, so half of each diagonal and one side form a right-angled triangle.

1. The diagonals bisect each other at right angles
2. So there is a right-angled triangle with hypotenuse 10 (a side) and one short side 5 (half the shorter diagonal)
3. Half the longer diagonal =\sqrt{10^2-5^2}=\sqrt{75}=5\sqrt3
4. (Or: the shorter diagonal cuts the rhombus into two equilateral triangles, and half the longer diagonal is 10\sin 60^\circ)
5. Longer diagonal =2\times5\sqrt3
6. Answer: 10\sqrt3 cm

Q and R have the same x-coordinate, so triangle PQR has a right angle at Q; with a 45^\circ angle it is isosceles.

1. PQ is horizontal and QR is vertical, so angle PQR=90^\circ
2. Angle QPR=45^\circ, so the triangle is isosceles: QR=PQ=5-(-3)=8
3. R is below Q (as k<2), so k=2-8=-6 and R is (5,\ -6)
4. Midpoint of PR: \left(\dfrac{-3+5}{2},\ \dfrac{2+(-6)}{2}\right)
5. Answer: (1,\ -2)

Draw a right-angled triangle with hypotenuse AB: the horizontal side is 8\cos 60^\circ and the vertical side is 8\sin 60^\circ.

1. Horizontal distance from A to B: 8\cos 60^\circ=8\times\dfrac12=4
2. Vertical distance from A to B: 8\sin 60^\circ=8\times\dfrac{\sqrt3}{2}=4\sqrt3
3. B is (1+4,\ 2+4\sqrt3)
4. Answer: (5,\ 2+4\sqrt3)

Look for a multiple of a Pythagorean triple you know, or use Pythagoras' theorem to find the length of the rectangle.

1. 2.1=7\times0.3 and 7.5=25\times0.3, so the sides are 0.3\times the triple 7,\ 24,\ 25
2. Length =24\times0.3=7.2 cm
3. Check: 7.5^2-2.1^2=56.25-4.41=51.84=7.2^2
4. Perimeter =2\times(2.1+7.2)
5. Answer: 18.6 cm

A regular hexagon splits into six equilateral triangles; find the height of one of them using \sin 60^\circ.

1. Each side is 24\div6=4 cm
2. The hexagon is made of six equilateral triangles of side 4 cm
3. Height of one triangle =4\sin 60^\circ=4\times\dfrac{\sqrt3}{2}=2\sqrt3
4. Area of one triangle =\dfrac12\times4\times2\sqrt3=4\sqrt3
5. Area of hexagon =6\times4\sqrt3
6. Answer: 24\sqrt3 cm^2

Write Pythagoras' theorem in full with the brackets, x^2+(x+7)^2=13^2, then expand and solve the quadratic.

1. Pythagoras: x^2+(x+7)^2=13^2
2. x^2+x^2+14x+49=169
3. 2x^2+14x-120=0, so x^2+7x-60=0
4. (x+12)(x-5)=0, so x=-12 or x=5
5. A length cannot be negative, so reject x=-12
6. Answer: x=5

Work out AC first, using triangle ABC and \cos 60^\circ; then use triangle ACD.

1. Triangle ABC: AB is adjacent to the 60^\circ angle and AC is the hypotenuse
2. \cos 60^\circ=\dfrac{6}{AC}, so AC=\dfrac{6}{\frac12}=12 cm
3. Triangle ACD: AC is adjacent to the 45^\circ angle and AD is the hypotenuse
4. \cos 45^\circ=\dfrac{12}{AD}, so AD=\dfrac{12}{\frac{1}{\sqrt2}}=12\sqrt2
5. Answer: 12\sqrt2 cm

Write \cos x^\circ in terms of h, then think about what happens to the angle as the hypotenuse h gets longer.

1. AB is adjacent to angle x and AC is the hypotenuse: \cos x^\circ=\dfrac{5}{h}
2. When h=5\sqrt2: \cos x^\circ=\dfrac{5}{5\sqrt2}=\dfrac{1}{\sqrt2}, so x=45
3. When h=10: \cos x^\circ=\dfrac{5}{10}=\dfrac12, so x=60
4. As h increases, \cos x^\circ decreases, so x increases
5. Answer: 45<x<60

AM cuts the triangle into two right-angled triangles, and BM is half of BC. Answer part (a) before attempting part (b).

Part (a)

1. AM is perpendicular to BC, and BM=\dfrac12(2x+2)=x+1
2. Pythagoras in triangle ABM: (x+1)^2+(x+3)^2=(x+5)^2
3. x^2+2x+1+x^2+6x+9=x^2+10x+25
4. Answer: x^2-2x-15=0

Part (b)

1. x^2-2x-15=0
2. (x-5)(x+3)=0, so x=5 or x=-3
3. x=-3 would make AM=0, so reject it
4. Answer: x=5

Part (c)

1. With x=5: AM=8 cm and AB=10 cm
2. In triangle ABM, AM is opposite angle ABC and AB is the hypotenuse
3. \sin ABC=\dfrac{8}{10}
4. Answer: \dfrac45

Triangle BCD is equilateral, and the diagonal AC cuts BD in half at right angles. Answer part (a) before attempting part (b).

Part (a)

1. Triangle ABD is an isosceles right-angled triangle
2. BD^2=6^2+6^2=72, so BD=\sqrt{72}=\sqrt{36}\times\sqrt2
3. Answer: 6\sqrt2 cm

Part (b)

1. AC crosses BD at right angles at its midpoint M, so BM=3\sqrt2
2. Triangle ABM has a 45^\circ angle at B, so AM=BM=3\sqrt2
3. CB=CD and angle BCD=60^\circ, so triangle BCD is equilateral with side 6\sqrt2
4. CM=6\sqrt2\times\sin 60^\circ=6\sqrt2\times\dfrac{\sqrt3}{2}=3\sqrt6
5. AC=AM+MC
6. Answer: 3\sqrt2+3\sqrt6 cm

Part (c)

1. Area of a kite =\dfrac12\times BD\times AC
2. =\dfrac12\times6\sqrt2\times(3\sqrt2+3\sqrt6)
3. =3\sqrt2\times3\sqrt2+3\sqrt2\times3\sqrt6=18+9\sqrt{12}
4. 9\sqrt{12}=9\times2\sqrt3=18\sqrt3
5. Answer: 18+18\sqrt3 cm^2

Angle ACB is an angle in a semicircle, so triangle ABC is right-angled at C. Answer part (a) before attempting part (b).

Part (a)

1. The angle in a semicircle is 90^\circ, so angle ACB=90^\circ
2. AC is opposite the 30^\circ angle and BC is adjacent: \tan 30^\circ=\dfrac{6}{BC}
3. BC=\dfrac{6}{\frac{1}{\sqrt3}}=6\sqrt3
4. Answer: 6\sqrt3 cm

Part (b)

1. \sin 30^\circ=\dfrac{6}{AB}, so AB=12 cm and the radius is 6 cm
2. Area of circle =\pi\times6^2=36\pi
3. Area of triangle ABC=\dfrac12\times AC\times BC=\dfrac12\times6\times6\sqrt3=18\sqrt3
4. Shaded area =36\pi-18\sqrt3
5. Answer: 36\pi-18\sqrt3 cm^2

Call AD h and write BD and DC in terms of h using the exact values for 60^\circ and 45^\circ. Answer part (a) before attempting part (b).

Part (a)

1. Let AD=h
2. Triangle ADC has a 45^\circ angle, so it is isosceles: DC=h
3. Triangle ADB: \tan 60^\circ=\dfrac{h}{BD}, so BD=\dfrac{h}{\sqrt3}
4. Area =\dfrac12\times BC\times AD=\dfrac12\left(h+\dfrac{h}{\sqrt3}\right)h
5. So \dfrac12 h^2\left(1+\dfrac{1}{\sqrt3}\right)=18+6\sqrt3
6. Multiply both sides by 2\sqrt3: h^2(\sqrt3+1)=36\sqrt3+36
7. h^2(\sqrt3+1)=36(\sqrt3+1), so h^2=36
8. Answer: AD=6 cm

Part (b)

1. Triangle ADC: AD=DC=6 and AC is the hypotenuse
2. AC=\dfrac{6}{\sin 45^\circ}=6\sqrt2 (or AC^2=6^2+6^2=72)
3. Answer: 6\sqrt2 cm

Drop a perpendicular from A to the x-axis and call its length h; write both parts of OC in terms of h.

Part (a)

1. Let D be the point on the x-axis directly below A, and let AD=h
2. Triangle AOD has a 45^\circ angle, so it is isosceles: OD=h
3. Triangle ACD: \tan 30^\circ=\dfrac{h}{DC}, so DC=\dfrac{h}{\tan 30^\circ}=h\sqrt3
4. OD+DC=12: h+h\sqrt3=12
5. h=\dfrac{12}{1+\sqrt3}
6. Rationalise: \dfrac{12}{1+\sqrt3}\times\dfrac{\sqrt3-1}{\sqrt3-1}=\dfrac{12(\sqrt3-1)}{3-1}=6\sqrt3-6
7. A is h across and h up from O
8. Answer: (6\sqrt3-6,\ 6\sqrt3-6)

Part (b)

1. Base OC=12 and perpendicular height = the y-coordinate of A=6\sqrt3-6
2. Area =\dfrac12\times12\times(6\sqrt3-6)
3. =6(6\sqrt3-6)
4. Answer: 36\sqrt3-36