10.2 Solving Problems Involving Angle of Elevation and Angle of Depression


Example:
The angle of depression of a cycling kid measured from a hill with 10.9 m high is 52o. When the kid cycles along the hillside and stops, the angle of depression becomes 25.3o. What is the distance cycled by the kid along the hillside?
 
Solution:


Step 1: Draw a diagram to represent the situation.
Step 2: Devise a plan.
Find the lengths of QS and QR. Then, QS – QR = distance cycled by the kid.
 

tan 52 o = 10.9 Q R Q R = 10.9 tan 52 o Q R = 8.5 m

tan 25.3 o = 10.9 Q S Q S = 10.9 tan 25.3 o Q S = 23.1 m

QS – QR = (23.1 – 8.5) m = 14.6 m

Therefore the kid has cycled 14.6 metres.

SPM Practice (Short Questions)


Question 1:
Diagram below shows two vertical poles on a horizontal plane. J, K, L, M and N are five points on the poles such that KL = MN.
Name the angle of elevation of point J from point M.
 
Solution:

Angle of elevation of point J from point M = ∠KMJ


Question 2:
Diagram below shows two vertical poles JM and KL, on a horizontal plane.


The angle of depression of vertex K from vertex J is 42o.
Calculate the angle of elevation of vertex K from M.

Solution:


JN = 10 tan 42o = 9.004 m
NM = 25 – 9.004 = 15.996 m
KL = NM = 15.996 m
tan K M L = K L M L = 15.996 10 = 1.5996 K M L = tan 1 1.5996 = 57 o 59 '



Question 3:
Diagram below shows a vertical pole KMN on a horizontal plane. The angle of elevation of from L is 20o.

Calculate the height, in m, of the pole.

Solution:


tan 20 o = K M K L 0.3640 = K M 15 K M = 5.4 m Height of the pole = 9 + 5.4 = 14.4 m

10.1 Angle of Elevation and Angle of Depression (Part 1)

10.1 Angle of Elevation and Angle of Depression (Part 1)

1. Angle of elevation is the acute angle measured from a horizontal line up to the line of sight.


Example 1:
Angle of elevation of object O from P = ∠OPA



2. Angle of depression is the acute angle measured down from a horizontal line to the line of sight.

 
Example 2:
Angle of depression of object O from P = ∠OPA

SPM Practice (Short Questions)


Question 4:
Diagram below shows three vertical poles JP, KN and LM, on a horizontal plane.


The angle of elevation of N from P is 15o.
The angle of depression of M from N is 35o.
Calculate the distance, in m, from K to L.
 
Solution:


tan A P N = A N P A tan 15 o = A N 4 A N = 4 × 0.268 A N = 1.072 m Length of B N = 2 + 1.072 = 3.072 c m tan B M N = B N B M tan 35 o = 3.072 B M B M = 3.072 0.700 = 4.389 Distance of K L = 4.389 m



Question 5:
Diagram below shows two vertical poles JM and LN, on a horizontal plane.

The angle of elevation of M from K is 70oand the angle of depression of K from N is 40o.
Find the difference in distance, in m, between JK and KL.
 
Solution:


tan J K M = 14 J K J K = 14 tan 70 o J K = 5.096 m tan L K N = 8 K L K L = 8 tan 40 o K L = 9.534 m Difference in distance of J K and K L = 9.534 5.096 = 4.438 m

9.2 Graphs of Sine, Cosine and Tangent

9.2 Graphs of Sine, Cosine and Tangent
 
(1)  y = sin x, 0o x ≤ 360o
 


x
0o
90o
180o
270o
360o
sin x
0
1
0
-1
0

 
(2)  y = cos x, 0o x ≤ 360o
 


x
0o
90o
180o
270o
360o
cos x
1
0
-1
0
1


 
(3)  y = tan x, 0o x ≤ 360o
 


x
0o
90o
180o
270o
360o
tan x
0
0
0



(4)  = sin 2x, 0≤ 2x ≤ 360o



(5)  = cos 2x, 0≤ 2x ≤ 360o




(6)  
= tan 2x, 0≤ 2x ≤ 360o
 


9.1 Values of Sine, Cosine and Tangent of an Angle (Part 1)

9.1 Values of Sine, Cosine and Tangent of an Angle

(A) Sine, Cosine and Tangent for Right-Angled Triangles
 

sin θ = opposites side hypotenuse cos θ = adjacent side hypotenuse tan θ = opposites side adjacent side

(B) Values of Sin θ, Cos θ and Tan θ in First Quadrant of the Unit Circle


In first quadrant, P is a point on the circumference of the unit circle with centre O, θ is the angle between the radius OP and the positive x-axis. From the diagram,
(a) sin θ = P Q O P = y (b) cos θ = O Q O P = x (c) tan θ = P Q O Q = y x T h e r e f o r e , sin θ = y -coordinate cos θ = x -coordinate tan θ = y -coordinate x -coordinate  
 
(C) A Cartesian plane is divided into four parts called quadrants by the x-axis and the y-axis. The quadrants are named as first quadrant, second quadrant, third quadrant and fourth quadrant in the anticlockwise direction.



Quick recall:
All – Add
Sine – Sugar
Tangent – To
Cosine – Coffee

Example:
Determine whether the value of each of the following is positive or negative.
(a) sin 105o (b) cos 75o (c) tan 305o (d) sin 50o   
(e) cos 160o (f) tan 220o (g) cos 260o (h) cos 350o

Solution:

(a) sin 105o is positive because 90o < 105o< 180o (in second quadrant).

(b) cos 75o is positive because 0o < 75o< 90o (in first quadrant).

(c) tan 305o is negative because 270o < 305o< 360o (in fourth quadrant).

(d) sin 50o is positive because 0o < 50o< 90o (in first quadrant).

(e) cos 160o is negative because 90o < 160< 180o (in second quadrant).

(f) tan 220o is positive because 180o < 220o< 270o (in third quadrant).

(g) cos 260o is negative because 180o < 260o< 270o (in third quadrant).

(h) cos 350o is positive because 270o < 350o< 360o (in fourth quadrant).

9.3 SPM Practice (Short Questions)


Question 4:



In the diagram above, WZY  is a straight line.  X Y Z = 90 o , X W Z = 30 o and WZ = XZ = 30 cm. Find the length of XY.

Solution:
WXZ=XWZ= 30 o XZY= 30 o + 30 o = 60 o sinXZY= XY XZ sin 60 o = XY 30 XY=sin 60 o ×30 XY=25.98cm


Question 5:



In the diagram above, PQS is a right angle triangle. Given that SR = 6cm, PQ = 12 cm and 5SR = 2PS. Find the value of cos α and tan β.

Solution:
5SR=2PS PS= 5 2 SR PS= 5 2 ( 6 ) PS=15 cm cosα= PQ PS cosα= 12 15 = 4 5 In  PQS, using Pythagoras' Theorem, QS= P S 2 P Q 2 QS= 15 2 12 2 =9 cm tanβ=tanPSQ Since  90 <β< 180 (in quadrant II), tanβ is negative tanβ= PQ QS tanβ= 12 9 = 4 3


Question 6:


In the diagram above, ADC is a straight line, if  sin q = 3 5 and tan p = 1 2 . Find the distance of AC.

Solution:
Given sin q = B D A B = 3 5 B D 30 = 3 5 B D = 3 5 × 30 B D = 18 c m In A B D , using Pythagoras' Theorem, A D = A B 2 B D 2 A D = 30 2 18 2 = 24 c m Given tan p = B D D C = 1 2 18 D C = 1 2 D C = 36 c m Hence, distance of A C = 24 + 36 = 60 c m .

9.1 Values of Sine, Cosine and Tangent of an Angle (Part 2)

(A) Special Angle






(B) Finding the Angles between 0oand 360o Given Values of sin θ, cos θ or tan θ
 
1. If the value of sin θ, cos θ or tan θ is given and 0o< θ < 360o, the value of θ can be found using the following steps.

(a) Find the basic angle in first quadrant which corresponds to θ.

(b) Based on the sign of the value of sin θ, cos θ or tan θ, determine the quadrants in which θ lies.

(c) Find the values of θ in the right quadrants found in (b).
 
Example:
For each of the following cases, find the value of θ.
(a) sin θ = 0.6025 and 90o < θ < 180o
(b) cos θ = –0.6025 and 180o < θ < 270o
(c) sin θ = –0.8387 and 0o < θ < 360o
(d) tan θ = –1.732 and 0o < θ < 360o
 
Solution:
(a)
sin θ = 0.6025
Basic ∠ = 37.05o ← (Press SHIFT sin-1 0.6025 =Display: 37.04976)
∴ θ = 180o – 37.05= 142.95o  ← (90o < θ < 180o)

(b)

cos θ = –0.6025
Basis ∠ = 52.95o ← (Press SHIFT cos-1 0.6025 = Display: 52.9508)
∴ θ = 180o + 52.95= 232.95o  ← (180o < θ < 270o)


(c)



sin θ = –0.8387 ← (sin θ is negative in 3rd quadrant and 4thquadrant)
Basis ∠ = 57o ← (Press SHIFT sin-1 0.8387 = Display: 57.003)
θ1 = 180o + 57o = 237o  ← (180o < θ < 270o)
θ2 = 360o – 57o = 303o  ← (270o < θ < 360o)
θ = 237o and 303o


(d)



tan θ = –1.732 ← (tan θ is negative in 2nd quadrant and 4thquadrant)
Basis ∠ = 60o ← (Press SHIFT tan-1 1.732 = Display: 59.9993)
θ1 = 180o – 60o = 120o  ← (90o < θ < 180o)
θ2 = 360o – 60o = 300o  ← (270o < θ < 360o)
θ = 120o and 300o


 

9.3 SPM Practice (Short Questions)


Question 1:


In the diagram above, find the value of tan θ.

Solution:

In A B C , using Pythagoras' Theorem, A C = 1 2 + 1 2 = 2 c m tan θ = C D A C tan θ = 1 2


Question 2:



In the diagram above, ABCE is a rectangle and point D lies on the straight line EC. Given that DC = 5 cm and AE = 4cm, find the value of cosθ.

Solution:
AD=DC=5cm In  AED, using Pythagoras' Theorem, ED= 5 2 4 2 =3cm cosθ=cosADE Since  90 <θ< 180 (in quadrant II), cosθ is negative cosθ= ED AD cosθ= 3 5



Question 3:



In the diagram above, PMR is a straight line, M is the midpoint of line PR. Given that QR = 12cm and sin y°= 0.6, find the value of tan x°.

Solution:
In triangle QMR sin y =0.6 sin y = QR QM = 6 10 Given QR=12cm, QM=10×2=20cm In  QMR, using Pythagoras' Theorem, MR= 20 2 12 2 =16cm PR=16×2=32cm Hence tan x = QR PR = 12 32 = 3 8