Higher Engineering Mathematics

(Greg DeLong) #1
126 GEOMETRY AND TRIGONOMETRY

Case 1.P= 46 ◦ 27 ′,Q= 36 ◦,R= 97 ◦ 33 ′,
p=36.5 mm andq= 29 .6 mm.
From the sine rule:

r
sin 97◦ 33 ′

=

29. 6
sin 36◦

from which,

r=

29 .6 sin 97◦ 33 ′
sin 36◦

=49.92 mm

Area=^12 pqsinR=^12 (36.5)(29.6) sin 97◦ 33 ′

=535.5 mm^2

Case 2.P= 133 ◦ 33 ′,Q= 36 ◦,R= 10 ◦ 27 ′,
p=36.5 mm andq= 29 .6 mm.
From the sine rule:

r
sin 10◦ 27 ′

=

29. 6
sin 36◦

from which,

r=

29 .6 sin 10◦ 27 ′
sin 36◦

=9.134 mm

Area=^12 pqsinR=^12 (36.5)(29.6) sin 10◦ 27 ′

=97.98 mm^2.

TrianglePQRfor case 2 is shown in Fig. 12.22.


Figure 12.22

Now try the following exercise.

Exercise 57 Further problems on solving
triangles and finding their areas

In Problems 1 and 2, use the sine rule to solve
the trianglesABCand find their areas.


  1. A= 29 ◦,B[= 68 ◦,b=27 mm.
    C= 83 ◦,a= 14 .1 mm,
    c= 28 .9 mm, area=189 mm^2


]


  1. B= 71 ◦ (^26) [′,C= 56 ◦ 32 ′,b= 8 .60 cm.
    A= 52 ◦ 2 ′,c= 7 .568 cm,
    a= 7 .152 cm, area= 25 .65 cm^2
    ]
    In Problems 3 and 4, use the sine rule to solve
    the trianglesDEFand find their areas.
    3.d=17 cm,[f=22 cm,F= 26 ◦.
    D= 19 ◦ 48 ′,E= 134 ◦ 12 ′,
    e= 36 .0 cm, area=134 cm^2
    ]
    4.d= 32 .6 mm,[ e= 25 .4 mm,D= 104 ◦ 22 ′.
    E= 49 ◦ 0 ′,F= 26 ◦ 38 ′,
    f= 15 .09 mm, area= 185 .6mm^2
    ]
    In Problems 5 and 6, use the sine rule to solve
    the trianglesJKLand find their areas.
    5.j= 3 .85 cm,k= 3 .23 cm,K= 36 ◦.



    J= 44 ◦ 29 ′,L= 99 ◦ 31 ′,
    l= 5 .420 cm, area= 6 .132 cm^2 or
    J= 135 ◦ 31 ′,L= 8 ◦ 29 ′,
    l= 0 .811 cm, area= 0 .916 cm^2



    6.k=⎡46 mm,l=36 mm,L= 35 ◦.


    K= 47 ◦ 8 ′,J= 97 ◦ 52 ′,
    j = 62 .2 mm, area= 820 .2mm^2 or
    K= 132 ◦ 52 ′,J= 12 ◦ 8 ′,
    j = 13 .19 mm, area= 174 .0mm^2



    12.10 Further worked problems on
    solving triangles and finding
    their areas
    Problem 23. Solve triangleDEFand find its
    area given thatEF= 35 .0 mm,DE= 25 .0mm
    and∠E= 64 ◦.
    TriangleDEFis shown in Fig. 12.23.
    Figure 12.23
    Applying the cosine rule:
    e^2 =d^2 + f^2 − 2 dfcosE
    i.e. e^2 =(35.0)^2 +(25.0)^2
    −[2(35.0)(25.0) cos 64◦]
    = 1225 + 625 − 767. 1 = 1083
    from which,e=

    1083 =32.91 mm

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