Page 172 - Mathematics_Form_Two

Trigonometry

2
2
( BC) =17 −8 2 Exercise 8�1
ˆ
1. In a triangle LMN, LNM 90= ° ,
2
=
BC = 17 − 8 2 LM = 10 cm, MN 6 cm, and
=
15cm= LN 8 cm. Find:
FOR ONLINE READING ONLY
(a) tan M (b) sin M (c) cos M
2. In a triangular plot ABC,
Mathematics for Secondary Schools Thus, DC =15cm−9cm and BC = 17 m. Find the value of
−
But, DC = BC - DBDC = BC - DB.
ˆ
BAC 90,= = 90º, AB = 8 m, AC = 15 m,
= 6 cm
each of the following:
BC
(a) sin C (b) tan C (c) cos C
Thus, cos x =
AC
ˆ
15 . 3. In a triangle RST, RST=90°,
cosx =
17 RS = 4cm, and TS = 3cm.. Find:
(a) TR (b) cos R (c) sin R
From ΔADC, apply Pythagoras theorem.
4. A rectangular field is 100 m long and
2
2
AD = 6 + 8 2 50 m wide. If one of its diagonals
makes an angle x with the length,
AD = 100 find the value of tan .x
=10 cm
5. Use the following figure to find:
(a) tan x
Thus, sin y = DC . (b) sin y
AD

6
sin y =
10

3
=
5

15 3
Hence, cos x + sin y = + . 4
17 5 6. If sin x = 5 , find the value of:
126 (a) tan x (b) cos x
=
85 15
126 7. If cos x = 17 , find the value of:
Therefore, cos x + sin y = .
85 (a) sin x (b) tan x

166
Student's Book Form Two

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MATHEMATIC F2 v5.indd 166