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Trigonometry
The longest side in a right-angled
triangle is called the hypotenuse (hyp).
xo
The side next to the angle is
called the adjacent (adj).
The side opposite
the angle
is called the
opposite (opp).
xo
hypotenuse
opposite
adjacent
tanx 
opposite
adjacent
cos x 
adjacent
hypotenuse
sinx 
opposite
hypotenuse
A useful memory aid for remembering these three ratios is:
SOHCAHTOA
Finding the length of a side
24o
x
6 cm
Example
1 Find the value of x.
opp
adj
tan24o

opp
adj
tan24o

x
6
x  6  tan24o
x  2.67 cm (to 3 s.f.)
replace opp by x and adj by 6
multiply both sides by 6
73o
x
8 cm
Example
2 Find the value of x.
adj
hyp
cos73o

adj
hyp
cos73o

x
8
x  8  cos73o
x  2.34 cm (to 3 s.f.)
replace adj by x and hyp by 8
multiply both sides by 8
40o
x 5 cm
Example
3 Find the value of x.
opp
hyp
sin40o

opp
hyp
sin40o

x
5
x  5  sin40o
x  3.21 cm (to 3 s.f.)
replace opp by x and hyp by 5
multiply both sides by 5
65o
x
15 cm
Example
4 Find the value of x.
adj
hyp
cos65o

adj
hyp
cos65o

15
x
x  cos65o
 15
x  35.5 cm (to 3 s.f.)
replace adj by 15 and hyp by x
multiply both sides by x
x 
15
cos65o
divide both sides by cos 65o
Finding an angle
xo
5 cm 9 cm
Example
1 Find the value of x.
opp
hyp
sinx 
opp
hyp
sinx 
5
9
  
  
 
1 5
sin
9
x
x  33.7o
(to 3 s.f.)
replace opp by 5 and hyp by 9
to find x use the sin-1 button on your calculator
xo
5 cm
8 cm
Example
2 Find the value of x.
adj
hyp
cos x 
adj
hyp
cos x 
5
8
  
  
 
1 5
cos
8
x
x  51.3o
(to 3 s.f.)
replace adj by 5 and hyp by 8
to find x use the cos-1 button on your calculator
xo
3 cm
6 cm
Example
3 Find the value of x.
opp
adj
tanx 
opp
adj
tanx 
3
6
  
  
 
1 3
tan
6
x
x  26.6o
(to 3 s.f.)
replace opp by 3 and adj by 6
to find x use the tan-1 button on your calculator
Three dimensional
trigonometry
To find the angle
between AG and the
base you need to
look at triangle AGC.
A B
7 cm
C
E
G
H
D
F
6 cm
5 cm
A B
7 cm
C
E
G
H
D
F
6 cm
5 cm
First you need to
calculate AC using
Pythagoras on
triangle ABC.
AC2
 72
 62
AC2
 85
 85
AC
85
Now use
trigonometry on
triangle ACG.

5
tan
85
x
  
  
 
1 5
tan
85
x
x  28.5o
(to 3 s.f.)
A C
G
5 cm
x

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Trigonometry for IGCSE Math Extended Book

  • 2. The longest side in a right-angled triangle is called the hypotenuse (hyp). xo The side next to the angle is called the adjacent (adj). The side opposite the angle is called the opposite (opp).
  • 3. xo hypotenuse opposite adjacent tanx  opposite adjacent cos x  adjacent hypotenuse sinx  opposite hypotenuse A useful memory aid for remembering these three ratios is: SOHCAHTOA
  • 4. Finding the length of a side
  • 5. 24o x 6 cm Example 1 Find the value of x. opp adj tan24o  opp adj tan24o  x 6 x  6  tan24o x  2.67 cm (to 3 s.f.) replace opp by x and adj by 6 multiply both sides by 6
  • 6. 73o x 8 cm Example 2 Find the value of x. adj hyp cos73o  adj hyp cos73o  x 8 x  8  cos73o x  2.34 cm (to 3 s.f.) replace adj by x and hyp by 8 multiply both sides by 8
  • 7. 40o x 5 cm Example 3 Find the value of x. opp hyp sin40o  opp hyp sin40o  x 5 x  5  sin40o x  3.21 cm (to 3 s.f.) replace opp by x and hyp by 5 multiply both sides by 5
  • 8. 65o x 15 cm Example 4 Find the value of x. adj hyp cos65o  adj hyp cos65o  15 x x  cos65o  15 x  35.5 cm (to 3 s.f.) replace adj by 15 and hyp by x multiply both sides by x x  15 cos65o divide both sides by cos 65o
  • 10. xo 5 cm 9 cm Example 1 Find the value of x. opp hyp sinx  opp hyp sinx  5 9         1 5 sin 9 x x  33.7o (to 3 s.f.) replace opp by 5 and hyp by 9 to find x use the sin-1 button on your calculator
  • 11. xo 5 cm 8 cm Example 2 Find the value of x. adj hyp cos x  adj hyp cos x  5 8         1 5 cos 8 x x  51.3o (to 3 s.f.) replace adj by 5 and hyp by 8 to find x use the cos-1 button on your calculator
  • 12. xo 3 cm 6 cm Example 3 Find the value of x. opp adj tanx  opp adj tanx  3 6         1 3 tan 6 x x  26.6o (to 3 s.f.) replace opp by 3 and adj by 6 to find x use the tan-1 button on your calculator
  • 14. To find the angle between AG and the base you need to look at triangle AGC. A B 7 cm C E G H D F 6 cm 5 cm A B 7 cm C E G H D F 6 cm 5 cm First you need to calculate AC using Pythagoras on triangle ABC. AC2  72  62 AC2  85  85 AC 85 Now use trigonometry on triangle ACG.  5 tan 85 x         1 5 tan 85 x x  28.5o (to 3 s.f.) A C G 5 cm x