Lesson Notes By Weeks and Term - Senior Secondary School 1
TRIGONOMETRIC RATIOS
TRIGONOMETRIC RATIOS
SUBJECT: MATHEMATICS
CLASS: SS 1
DATE:
TERM: 2nd TERM
REFERENCE BOOK
WEEK SEVEN
TOPIC: TRIGONOMETRIC RATIOS
CONTENT
Sine, Cosine and Tangent of acute angles
Given a right angled triangle, the trigonometric ratio of acute angles can be found as shown below
In the figure above, ABC is any triangle, right-angled at A
tan B = b tan C = c(tan : Opp)
c b Adj
Sin B = b Sin C =c Sin ;Opp
aaHyp
Cos B = c Cos C = b ( Cos : Adj)
aaHyp
InABC, B and C are complementary angles i.e B + C = 90o
If B = Ó¨ then C = 90o - Ó¨.
In the figure above sin Ó¨ =cos ( 90-Ó¨ ) = b
a
Cos Ó¨ = Sin (90- Ó¨) = c
a
Note: Always remember SOH CAH TOA
i.e Sin θ = Opp
Hyp
Cos θ= Adj
Hyp
Tan O =Opp
Adj.
Examples
Solutions
(1)
tan θ = 8cm = 1.600
5cm
Ó¨ = tan-1 1.6000
Ó¨ = 58o.
The 3rd angle in the right angled triangle above = 90o – 58o = 32.
Hence, the smallest angle of the given triangle = 32o
Y
2)
From the diagram drawn above the distance of Y east of X = ZY
Using the right angled triangle XZY
Sin 40o =ZY
200km
ZY = 200km x sin 40o
= 200km x 0.6428
= 128.56km
= 128.6km to 1d.p
3)
In the given diagram,
LK = Sin 70o
7cm
LK = 7cm x sin 70o ………… i
LK = 7cm x 0.9397
LK = 6.5779cm
In right angled triangle LKN
LK = sin MNL
21
i.e 7cm x 0.9397 = Sin MNL
21cm
3
0.3132 = Sin MNL
Sin-10.3132 = MNL
i.e MNL = 18.3o
EVALUATION
A ladder 20cm long rests against a vertical wall so that the foot of the ladder is 9m from the wall.
(a) Find, correct to the nearest degree, the angle that the ladder makes with the wall
(b) Find correct to 1.dp the height above the ground at which the upper end of the ladder touches the wall. Use of tables of trigonometric ratios.
Determination of lengths of chords using trigonometric ratios
Trignometric ratios can be used to find the length of chords of a given circle. However, in some cases where angles are not given, Pythagoras theorem is used to find the lengths of chords in such cases. Pythagoras theorem is stated as follows:
It states that c2 = a2 + b2
Pythagoras theorem states that in a right angled triangle, the square of the length of the hypotenuse is equal to the sum of the square of the lengths of the other two sides.
Examples
58o
Solutions
1
B
From the diagram above in right-angled triangle ABO:
AB 2 + 32 = 52 ( Pythagoras theorem)
AB 2= 52 - 32
AB 2 = 25 – 9
AB 2 = 16
AB = √16 = 4cm
Since B is the mid point of chord AC then
Length of chord AC = 2 x AB
= 2x 4cm =8cm
2)
0
Let the distance from O to HL= xcm
In right-angled triangle OMH:
OH 2 = HM 2 + MO 2
OH 2= 82 + 42
= 64 + 16
= 80
:. OH = √80
:. OH = √80cm
but OH = radius of the circle
i.e r= OH = OL = √80cm
In right-angled triangle ONL
OL 2 = ON 2 + NL 2
i.e( √80)2 = x2 + 52
80- 25 = x2
55 = x2
Take square root of both sides
√55 = √x2
√55 = x = 7. 416cm
:. The length of the perpendicular from O to HL is 7.416cm
3)
The perpendicular from O to AB divides the vertical angle into 2 equal parts and also divides the length of chord AB into two equal parts.
In right-angled triangle ACO:
AC = Sin 29o
OA 1
Cross multiply
AC = OA x sin 29o
AC = 14cm x Sin 29o
AC = 14cm x 0. 4848
AC = 6.787cm
AB = 2 x 6.787 cm
AB = 13.574cm
:. The length of the chord AB = 13.6cm to 1 d.p
EVALUATION
GRAPH OF SINE AND COSINE FOR ANGLES
In the figure below, a circle has been drawn on a Cartesian plane so that its radius, OP, is of length 1unit. Such a circle is called unit circle.
The angle Ѳ that OP makes with Ox changes according to the position of P on the circumference of the unit circle. Since P is the point (x,y) and /OP/ = 1 unit,
Sin Ѳ = y/1 = y
Cos Ѳ = x/1 = x
Hence the values of x and y give a measure of cos Ѳ and sin Ѳ respectively.
If the values of Ѳ are taken from the unit circle, they can be used to draw the graph of sin Ѳ. This is done by plotting values of y against corresponding values of Ѳ as in the figure below.
In the figure above, the vertical dotted lines gives the values of sin Ѳ corresponding to Ѳ = 30o, 60 o,
90 o,......., 360 o.
To draw the graph of cosѲ , use corresponding values of x and Ѳ. This gives another wave-shaped curve, the graph of cos Ѳ as in the figure below.
As Ѳ increases beyond 360o, both curves begin to repeat themselves as in the figures below.
Take note of the following:
1)All values of sin Ѳ and cos Ѳ lie between +1 and -1.
2)The sine and cosine curves have the same shapes but different starting points.
3)Each curve is symmetrical about its peak(high point) and trough(low point). This means that for any value of sin Ѳ there are usually two angles between 0 o and 360 o; likewise for cos Ѳ. The only exceptions to this are at the quarter turns, where sinѲ and cosѲ have the values given in the table below
0o | 90o | 180o | 270o | 360o | |
SinѲ | 0 | 1 | 0 | -1 | 0 |
CosѲ | 1 | 0 | -1 | 0 | 1 |
Examples
1) Referring to graph on page 194 0f NGM Book 1, a)Find the value of sin 252o, b)solve the equation 5 sin Ѳ = 4
Solution
a)On the Ѳ axis, each small square represents 6. From construction a) on the graph:
Sin 252o = -0.95
b)If 5 sin Ѳ = 4
then sin Ѳ = 4/5 = 0.8
From construction b) on the graph: when sin Ѳ = 0.8, Ѳ = 54o or 126o
EVALUATION
1)Using the same graph used in the above example, find the values of the following
a)sin 24o b) sin 294o
2)Use the same graph to find the angles whose sines are as follows:
GENERAL EVALUATION
READING ASSIGNMENT
NGM SS BK 1 pg 114- 123, Ex 11a .
NOS 10 and 25 pg 117 -118
WEEKEND ASSIGNMENT
THEORY
© Lesson Notes All Rights Reserved 2023