# Lesson Notes By Weeks and Term - Senior Secondary School 2

BEARINGS AND DISTANCES

SUBJECT: MATHEMATICS

CLASS:  SS 2

DATE:

TERM: 3rd TERM

REFERENCE BOOKS

• New General Mathematics SSS2 by M.F. Macrae etal.
• Essential Mathematics SSS2 by A.J.S. Oluwasanmi.
• Exam Focus Mathematics.

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

Ó¨

C     B

Angle ACB = Ó¨ = 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 = Ó¨ = Angle of Depression.           B       Ó¨

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 â 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.

40

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 â BDC, Tan Ó¨ = 1.54/20

Ó¨ = 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 âACB, Tan 430 = AB/x

AB = x tan 430     ……………………eqn 1

From â 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

1. 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 = 075

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 â )

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 0090.

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.

1. 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

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.S600
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.