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LAW OF SINES TO FIND MISSING SIDE

The sine rule is a set of equations which connects the lengths of the sides of any triangle with the sine of the opposite angles.

The triangle does not have to be right angled for sine rule to be used.

In any triangle ABC, with sides a, b and c units in length and opposite sides A, B and C respectively.

asin A = bsin B = csin C(or)sin Aa = sin Bb = sin Cc
sine-law

Solve for the unknown in each triangle. Round to the nearest tenth.

Problem 1 :

sine-law-q1

Solution:

From the picture given above, we know that we need to find the angle A.

BC = a = 22 m, AB = c = 17 m and AC = b 

∠C = 42°

asin A=bsin B=csin C22sin A=17sin 42°22sin A=170.66922sin A=25.41sin A=2225.41sin A=0.865A=59.88°

Problem 2 :

sine-law-q2

Solution:

AC = b = 44 mm, AB = c = x, BC = a

∠A = 35°, ∠B = 88°

Here, 

∠C = 180° - (35° + 88°)

∠C = 57°

asin A=bsin B=csin C44sin 88°=xsin 57°x(sin 88°)=44(sin 57°)x=44(sin 57°)sin 88°x=44×0.8380.999x=36.90 mm

Problem 3 :

sine-law-q3

Solution:

AB = c = 9.4 cm, BC = a = 6 cm, AC = b

∠C = 51°

asin A=bsin B=csin C6sin A=9.4sin 51°6sin A=9.40.777sin A(9.4)=6(0.777)sin A(9.4)=4.662sin A=4.6629.4sin A=0.495A=29.669°

Problem 4 :

sine-law-q4

Solution:

AC = b = 12 m, AB = c = 13 m, BC = a

∠B = 67° 

asin A=bsin B=csin C12sin 67°=13sin C120.92=13sin C12 sin C=13(0.92)12 sin C=11.96sin C=11.9612sin C=0.996C=84.87°

Problem 5 :

sine-law-q5

Solution:

AB = c = x, BC = a = 21 cm, AC = b

∠B = 48°, ∠C = 61°

∠A = 180° - (48° + 61°)

∠A = 71°

asin A=bsin B=csin C21sin 71°=xsin 61°210.945=x0.8740.945x = 21×0.8740.945x=18.354x=18.3540.945x=19.42 cm

Problem 6 :

sine-law-q6

Solution:

AB = c = 45 m, AC = b = x, BC = a

∠A = 118°, ∠C = 52°

∠B = 180° - (118° + 52°)

∠B = 10°

asin A=bsin B=csin Cxsin 10°=45sin 52°x0.173=450.7880.788x = 45×0.1730.788x=7.785x=7.7850.788x=9.92 m
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