Everything Maths Grade 11

(Marvins-Underground-K-12) #1

17.3 CHAPTER 17. TRIGONOMETRY



A


B


C


45 ◦


Figure 17.7: An isosceles right angled triangle.

Take any right-angled triangle with one angle 45 ◦. Then, because one angle is 90 ◦, the third angle is
also 45 ◦. So we have an isosceles right-angled triangle asshown in Figure 17.7.


If the two equal sides are of length a, then the hypotenuse, h, can be calculated as:


h^2 = a^2 + a^2
= 2a^2
∴ h =


2 a

So, we have:


sin(45◦) =
opposite(45◦)
hypotenuse
=
a

2 a

=

1



2


cos(45◦) =

adjacent(45◦)
hypotenuse
=
a

2 a
=

1



2


tan(45◦) =
opposite(45◦)
adjacent(45◦)
=
a
a
= 1

We can try something similar for 30 ◦and 60 ◦. We start with an equilateral triangle and we bisect one
angle as shown in Figure 17.8. This gives us theright-angled triangle that we need, with one angle of
30 ◦and one angle of 60 ◦.


If the equal sides are oflength a, then the base is^12 a and the length of the vertical side, v, can be
calculated as:


v^2 = a^2 − (

1


2


a)^2

= a^2 −

1


4


a^2

=

3


4


a^2

∴ v =


3


2


a
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