Higher Engineering Mathematics

INTRODUCTION TO TRIGONOMETRY 119

B

Figure 12.11

hence XZ= 20 .0 sin 23◦ 17 ′

= 20 .0(0.3953)=7.906 mm

cos 23◦ 17 ′=

XY

20. 0

hence XY= 20 .0 cos 23◦ 17 ′

= 20 .0(0.9186)=18.37 mm

[Check: Using Pythagoras’ theorem

(18.37)^2 +(7.906)^2 = 400. 0 =(20.0)^2 ]

Area of triangleXYZ

=^12 (base) (perpendicular height)

=^12 (XY)(XZ)=^12 (18.37)(7.906)

=72.62 mm^2

Now try the following exercise.

Exercise 54 Further problems on the solu-

tion of right-angled triangles

1. Solve triangleABCin Fig. 12.12(i).
   [
   BC= 3 .50 cm,AB= 6 .10 cm,
   ∠B= 55 ◦

]

Figure 12.12

2. Solve triangleDEFin Fig. 12.12(ii)
   [FE=5 cm,∠E= 53 ◦ 8 ′,∠F= 36 ◦ 52 ′]


3. Solve triangle[ GHIin Fig. 12.12(iii)
   GH= 9 .841 mm,GI= 11 .32 mm,
   ∠H= 49 ◦

]

4. Solve the triangleJKLin Fig. 12.13(i) and

find its area

[

KL= 5 .43 cm,JL= 8 .62 cm,

∠J= 39 ◦, area= 18 .19 cm^2

]

5. Solve the triangleMNOin Fig. 12.13(ii) and
   find its area[
   MN= 28 .86 mm,NO= 13 .82 mm,
   ∠O= 64 ◦ 25 ′, area= 199 .4mm^2

]

Figure 12.13

6. Solve the trianglePQRin Fig. 12.13(iii) and
   find its area[
   PR= 7 .934 m,∠Q= 65 ◦ 3 ′,
   ∠R= 24 ◦ 57 ′, area = 14 .64 m^2

]

7. A ladder rests against the top of the perpen-
   dicular wall of a building and makes an angle
   of 73◦with the ground. If the foot of the lad-
   der is 2 m from the wall, calculate the height
   of the building. [6.54 m]

12.5 Angles of elevation and depression

(a) If, in Fig. 12.14, BC represents horizontal

ground and ABa vertical flagpole, then the

angle of elevationof the top of the flagpole,

A, from the pointCis the angle that the imagi-

nary straight lineACmust be raised (or elevated)

from the horizontalCB, i.e. angleθ.

Figure 12.14

(b) If, in Fig. 12.15,PQrepresents a vertical cliff

andRa ship at sea, then theangle of depression

of the ship from pointPis the angle through

which the imaginary straight linePRmust be

lowered (or depressed) from the horizontal to

the ship, i.e. angleφ.

Figure 12.15