Week 3 – SS2 Third Term Mathematics Notes

WEEK THREE
TOPIC: BEARINGS AND DISTANCES

• Bearings And Distances
  
• Angles of Elevation and Depression
  

Angle of Elevation
This is the angle formed between the normal eye level and the line through which the observer view an object above.
A

 
 
 
 
  Ө
C B
Angle ACB = Ө = Angle of elevation.

 Angle of Depression: This is the angle formed between the eye level of the observer and the object below when the observer is above the object at view.
Angle ABC = Ө = Angle of Depression. B Ө

 
 
 
 AC
The angle of elevation is alternate to the angle of depression and problems involving angles of elevation and depression could be solved by using the basic trigonometric ratios and in some cases the sine and cosine rule could be applied.
Sine Rule for ∆ ABC; a = b = c
Sin A Sin B Sin C
Cosine Rule: a² = b² + c² – 2bc Cos A
b² = a² + c² – 2ac Cos B
c² = a² + b² – 2ab Cos C
EXAMPLES:

1. A ladder 50m long rests against a vertical wall. If the ladder makes an angle of 65⁰ with the ground, find the distance between the foot of the ladder and the wall.

Solution: Q Ladder = QR, Wall = QP
Distance between the foot of the ladder and the wall = PR

  50m Cos 65⁰ = PR
50
Cross multiplying
PR = 50 x Cos 65⁰ = 50 x0.4226
PR = 21.13m

P 65⁰ R

1. The angle of depression of an object on the ground from the top of a tower 60m high is 55⁰. Find the distance between the foot of the tower and the object to the nearest whole number.

Solution: A
55⁰ Tower = AC, Object = A
Distance between the foot of the tower and
     the object = BC

  Tan 55⁰ = AC, tan 55⁰ = 60
60m BC BC
Cross multiplying; BC x Tan 55⁰ = 60
BC = 60/tan 55⁰
BC BC = 60/1.428 = 42.02m
BC = 42m (nearest whole number)
EVALUATION
From the top of a building 10m high, the angle of elevation of a stone lying on the horizontal ground is 70⁰. Calculate correct to 1 decimal place, the distance of the stone from the foot of the building and the distance of the stone from the top of the building.

 FURTHER EXAMPLES:

1. The angle of elevation of the top of a vertical pole from a height 1.54m above a horizontal ground is 40⁰. The foot of the pole is on the same horizontal ground and the point of observation is 20m from the pole. Calculate correct to 3 significant figures. (i) the height of the pole (ii) the angle of depression of the foot of the pole from the point of observation.

Solution:E
Pole = BE, AD = Point above the ground.
40⁰
D C
1.54m

 
A     B
20m

1. BE = Height of the pole = BC + CE

BC = AD = 1.54m, AB = CD = 20m (opposite & parallel sides).
To obtain CE, using triangle CDE;
Tan 40⁰ = CE/CD, Tan 40⁰ = CE/20
Cross multiplying; CE = 20 x tan 40⁰ = 20 x 0.8391
CE = 16.782m
Therefore, Height of the pole (BE) = 1.54 + 16.782 = 18.322m
= 18.3m (3 sig. fig)

1. Angle of depression of the foot of the pole from point D:

   Using ∆ BDC, Tan Ө = 1.54/20
   Ө = tan^-1(0.077) = 4.41⁰

1. A boy observes that the angle of elevation of the top of a tower is 32⁰. He then walks 8m towards the tower and then discovers that the angle of elevation is 43⁰. Find the height of the tower to the nearest metre.

   Solution:
   A
   
    
    
    
    
     32⁰ 43⁰
   D 8m C x B
   Height of the tower is AB, using ∆ACB, Tan 43⁰ = AB/x
   AB = x tan 43⁰ ……………………eqn 1
   From ∆ ADB, Tan 32⁰ = AB/(8 + x)
   AB = (8 + x) tan 32⁰ ……………………eqn 2
   Equating the two equations: x tan 43⁰ = (8 + x) tan 32⁰
   x tan 43⁰ = 8tan 32⁰ + x tan 32⁰
   x tan 43⁰ – x tan 32⁰ = 8tan 32⁰
   x(tan 43⁰ – tan 32⁰) = 8 tan 32⁰
   x = 8 tan 32⁰
   tan 43⁰ – tan 32⁰
   x = 8 x 0.6249 = 4.9992
   0.9325 – 0.6249 0.3076
   x = 16.252m
   Height of the tower = AB = x tan 43⁰ = 16.252 x 0.9325 = 15.15499m
   Height of the tower = 15.2m

    

EVALUATION
The feet of two vertical pole of height 3m and 7m are in line with a point P on the ground, the smaller pole being between the taller pole and P and at a distance of 20m from P. The angle of elevation of the top (T) of the taller pole from the top (R) of the smaller pole is 30⁰. Calculate:

1. Distance RT (b) Distance of the foot of the taller pole from P, correct to 3 significant figures.
2. Angle of elevation of T from P, correct to one decimal place.

 BEARING AND DISTANCES
Bearings can be defined as the angular relationships between two or more places. Bearings are specified in two ways:

 Cardinal Points: It is specified in reference to the north and south. E.g N45⁰E, S60⁰W

 
 
 
 
 
 
 
 
 
 Example
Taking O as the starting point.
NOP = 60⁰ is the bearing N60⁰E
SOQ= S47⁰W

 
 Three Digits Notation: Bearing is also specified in three digits notation. E.g 060⁰, 078⁰,135⁰,225⁰ e.t.c

 Example1. Indicate the following bearing on the cardinal point (a) 080⁰ (b) 210⁰
Solution
(a) 080⁰, ( b) 210⁰




 

 2. Write each of the following in three digit notation. (a) S 70⁰E (b) N40⁰W
Solution:
(a)S 70⁰E

 
 
 
  It is equivalent to 110⁰
(b) N 40⁰W

 
 
 
 
 It is equivalent to 320⁰

 Evaluation
Find the equivalent of the following in three digit notation. 1. S 75⁰W 2. N 35⁰E 3.S 30⁰E 4.N62⁰W

 Bearing of One Point from Another;
It is possible to determine the bearing of one point or location from another point, if the starting point is known.

 Examples
1.Find the bearing of A from B if B from A is 140⁰.
Solution;
A from B = ? B from A = 140⁰

 A from B =270⁰ + 50⁰ =320⁰
2.If the bearing of P from Q is 075⁰, find the bearing of Q from P.
Solution;
P from Q = 075⁰

R from P = 180 + 75
= 255⁰

 
 Evaluation: Find the bearing of X from Y, if Y from X is 210⁰.

 The Sine rule and Cosine rule are the basic rule used to solve bearing related problems.
Sine rule; Sin A = Sin B = Sin C or a = b = c
a b c Sin A Sin B Sin C
Cosine rule; c² = a² + b² – 2abCosC
a² = b² + c² – 2bcCosA
b² = a² + c² – 2acCosB
Examples
1. A fly moves from a point U on a bearing of 060⁰, to a point V 20m away. It then moves from the point V on a bearing of 130⁰, to a point W. If the point W is due east of U. Find the distance of the point V from W and U from W.
Solution

 
 U + V + W = 180⁰ ( sum of angles in a ∆ )
W = 180⁰ – 30⁰ – 110⁰, W = 40⁰
To find the distance V from W, using sine rule; Sin U = Sin W
u w
sin 30⁰ = sin 40⁰
u 20, u = 20 sin 30⁰
sin 40⁰
u = 15.56mdistance V from W = 15.6m (3 s.f)
Distance u from w; sin U = sin V
u v
sin 30 = sin 110⁰
15.6 v

  v = 15.6 x sin 110⁰ v = 29.32
sin 30
hence, distance of u from w = 29.3m (3 s.f).

 2.A village R is 10km from a point P on a bearing 025⁰ from P. Another village A is 6km from P on a bearing 162⁰. Calculate (a) distance of R from A (b)the bearing of R from A.
Solution:

 (a) Distance R from A, using cosine rule: p² = q² + r² – 2qr Cos P
P² = 10² + 6² – 2(10 x 6) Cos 137⁰
P² = 100 + 36 – (120) x( -0.7314)
P² = 136 + 87.768
P = √223.768, p = 14.96km
Distance R from Q = 15km approximately.

 (b)Bearing of R from Q, Let the bearing be x, to find x, find A first

 Sin Q = Sin P
a p
Sin A = Sin 137⁰
10 14.96     
Sin Q = 10 x Sin137⁰
14.96
Q = sin^-10.4559, Q = 27.1⁰
But, Q = 18 + x
27.10=18+x
x=27.10 – 18=09.1⁰
The bearing of R from Q is 009^0.

 Evaluation
City A is 300km due east of city B.City C is 200km on a bearing of 1230 from city B.How far is it from C to A?

 General Evaluation:
1)Find the corresponding bearing of the following: (a)N64⁰W (b)064⁰ (c)S42⁰E (d) 234⁰
2)If the bearing of X from Y is N64⁰W.Find the bearing of Y from X.
3)A boat sails 6km from a port X on a bearing of 065⁰ and thereafter 13km on a bearing of 136⁰. What is the distance and bearing of the boat from X.
4. Find the angle of elevation to the nearest degree of the top of a church tower 180m high from a point on the ground 75m from its foot.

 Revision Questions
1 From a place 400m north of X, a student walks eastward to a place Y which is 800m from X. What is the bearing of X from Y
2 In a circle of radius 18cm, two radii form an angle of 150⁰ at the centre from point X and Y on the circumference. Find correct to three significant figure
(a) the length of the chord XY
(b) the length of the major arc
(c) the area of the minor segment

 Reading Assignment
Essential Mathematics SSS2, pages 195-197, nos 1-10.

 Weekend Assignment
Objectives
1. What is the equivalent of S70⁰E in three digit notation? A. 110⁰ B. 070⁰ C.120⁰ D.100⁰
2. If the bearing of P from R is 065⁰, what is R from P? A. 230⁰ B.245⁰ C. 120⁰ D 025⁰
3. Express the true bearing 210⁰ as a compass bearing. A S30⁰W B S60⁰W C.N30⁰W D.S60⁰E
4. Town Q is on a bearing 210⁰ from town P, town R is on a bearing 150⁰ from town P and R is east of Q. The distance between R and P is 10km. Find the distance between R and Q. A. 10km B. 20km C.30km D. 40km
5. What is the bearing of M from N, if the bearing of N from M is 315⁰? A.065⁰ B. 015⁰ C. 045⁰ D. 025⁰

 Theory
1. P, Q and R are points in the same horizontal plane. The bearing of Q from P is 150⁰ and the bearing of R from Q is 060⁰. If |PQ| = 5m and |QR| = 3m. Find the bearing of R from P correct to the nearest degree.
2.The angles of elevation of the top T, of a tower, 25m high are observed from point A at the top of a building to be 38⁰ and from point B at the bottom of the building to be 65.4⁰. If the tower and the building are on the same horizontal level, calculate (a) BT (B) the height of the building. Give your answers correct to 3 s.f.