Page 176 - Mathematics_Form_Two

Trigonometry

Example 8�8 12
= = 62
Find the exact value of 2sin 60°+ cos30 .° 2
Therefore, RQ = 6 2 cm.
Solution
Using the values in Table 8.1, it follows Example 8�10
that; 0 2 3 and cos 30° = 2 3 � Find the values of a, b, and c in the
FOR ONLINE READING ONLY
Mathematics for Secondary Schools Thus, o cos30 o = 33 2 3 + 2 3
0
sin 60
cos30
sin 60° =
triangle LMP.
2sin 60° + cos 30°= 2×

2cos60 +
.
=
2
33
Therefore, 2cos 60° + cos 30° =
2cos60 +
o
o
cos30
2 .
Example 8�9
Solution
An isosceles triangle PQR is such In ∆LMN, apply the Pythagoras'
ˆ
ˆ
that PQR = 45° and RPQ 90 .= ° If theorem. That is,
2
2
PQ = 6cm, find the length of RQ, a = 10 + 10 2
2
giving the answer in radical form. a = 200
Solution a =10 2 cm
Consider the triangle PQR. In ∆LMP, it implies that,

LM 10
tan30° = =
MP MP

Thus, 3 = 10 .
3 MP

10 3
×
MP =
From the ΔPQR,it follows that 3
6 10 3= cm
cos 45° =
RQ
But MP 10 b= + .
6 6
RQ = =
cos 45° 2 10 3 10 b= +
6 6 2 .
RQ = = Thus, b = 10 ( 3 1 cm− )
cos 45° 2
2 In ∆LMN, it follows that,

170
Student's Book Form Two

11/10/2024 20:13:52
MATHEMATIC F2 v5.indd 170
MATHEMATIC F2 v5.indd 170 11/10/2024 20:13:52