FINDING THE MISSING SIDES AND ANGLES USING LAW OF SINES

Problem 1 :

Find AC

missing-sides-angles-using-law-of-sines-q1

Solution :

CB = a, AB = c = 24, AC = b

∠̇A = 118º, ∠̇B = 22º

In triangle ABC, 

∠̇A + ∠̇B + ∠̇C = 180º

118º + 22º + ∠̇C = 180º

140º + ∠̇C = 180º

∠̇C = 180º - 140º

∠̇C = 40º

asin A = bsin B = csin C CBsin 118° = ACsin 22° = ABsin 40° CBsin 118° = ACsin 22° = 24sin 40° ACsin 22° = 24sin 40°AC0.375 = 240.643AC0.375= 37.33AC = 37.33 × 0.375AC = 13.9AC = 14

So, the missing side and missing angle are 14 and 40 degree.

Problem 2 :

Find AB

missing-sides-angles-using-law-of-sines-q2

Solution :

CB = a, AC = b = 7, AB = c

∠̇B = 44º, ∠̇C = 53º

In triangle ABC, 

∠̇A + ∠̇B + ∠̇C = 180º

∠̇A + 44º + 53º = 180º

∠̇A + 97º = 180º

∠̇A = 180º - 97º

∠̇A = 83º

asin A = bsin B = csin C CBsin 83° = ACsin 44° = ABsin 53° CBsin 83° = 7sin 44° = ABsin 53° 7sin 44° = ABsin 53°70.695 = ABsin 53°10.072= AB0.799AB = 10.072 × 0.799AB = 8

So, the missing side and missing angle are 8 and 83 degree.

Problem 3 :

Find BC

missing-sides-angles-using-law-of-sines-q3

Solution :

CB = a, AC = b = 27, AB = c

∠̇A = 39º, ∠̇C = 51º

In triangle ABC, 

∠̇A + ∠̇B + ∠̇C = 180º

∠̇39º + ∠̇B + 51º = 180º

∠̇B + 90º = 180º

∠̇B = 180º - 90º

∠̇B = 90º

asin A = bsin B = csin C CBsin 39° = ACsin B = ABsin 51° CBsin 39° = 27sin 90° = ABsin 51° CBsin 39° = 27sin 90° CB0.629 = 271CB = 27 × 0.629CB = 16.983CB = 17

So, the missing side and missing angle are 17 and 90 degree.

Problem 4 :

Find AB

missing-sides-angles-using-law-of-sines-q4

Solution :

CB = a = 9, AC = b, AB = c

∠̇B = 101º, ∠̇C = 63º, 

In triangle ABC, 

∠̇A + ∠̇B + ∠̇C = 180º

∠̇A + 101º + 63º = 180º

∠̇A + 164º = 180º

∠̇A = 180º - 164º

∠̇A = 16º

asin A = bsin B = csin C CBsin 16° = ACsin 101° = ABsin 63° 9sin 16° = ACsin 101° = ABsin 63° 9sin 16° = ABsin 63° 90.276 = AB0.89132.61 = AB0.891AB = 32.61 × 0.891AB = 29.05AB = 29.1

So, the missing side and missing angle are 29.1 and 16 degree.

Problem 5 :

Find BC

missing-sides-angles-using-law-of-sines-q5

Solution :

CB = a, AC = b = 16, AB = c

∠̇A = 93º, ∠̇C = 58º

In triangle ABC, 

∠̇A + ∠̇B + ∠̇C = 180º

93º + ∠̇B  + 58º = 180º

∠̇B + 151º = 180º

∠̇B = 180º - 151º

∠̇B = 29º

asin A = bsin B = csin C CBsin 93° = ACsin 29° = ABsin 58° CBsin 93° = 16sin 29° = ABsin 58° CBsin 93° = 16sin 29° CB0.999 = 160.485CB0.999 = 32.99CB = 32.99 × 0.999CB = 32.96CB = 33

So, the missing side and missing angle are 33 and 29 degree.

Problem 6 :

Find m∠̇C

missing-sides-angles-using-law-of-sines-q6

Solution :

CB = a = 26, AC = b = 16.1, AB = c = 21

∠̇A = 88º 

asin A = bsin B = csin C 26sin 88° = 16.1sin B = 21sin C 26sin 88° = 21sin C 260.999 = 21sin C26 sin C = 21 × 0.99926 sin C = 20.98sin C = 20.9826sin C = 0.807C = sin -1 (0.807)C =53.8C = 54°

Hence, the required angle is m∠̇C = 54º.

Problem 7 :

Find m∠̇C

missing-sides-angles-using-law-of-sines-q7

Solution :

CB = a = 29, AC = b = 24, AB = c = 20

∠̇A = 82º 

asin A = bsin B = csin C 29sin 82° = 24sin B = 20sin C 29sin 82° = 20sin C 290.99 = 20sin C29 sin C = 20 × 0.9929 sin C = 19.8sin C = 19.829sin C = 0.683C = sin -1 (0.683)C =43.07C = 43.1°

Hence, the required angle is m∠̇C = 43.1º.

Problem 8 :

Find m∠̇C

missing-sides-angles-using-law-of-sines-q8

Solution :

CB = a = 26, AC = b = 24, AB = c = 6

∠̇A = 103º 

asin A = bsin B = csin C 26sin 103° = 24sin B = 6sin C 26sin 103° = 6sin C 260.974 = 6sin C26 sin C = 6 × 0.97426 sin C = 5.844sin C = 5.84426sin C = 0.223C = sin -1 (0.223)C =12.88C = 13°

Hence, the required angle is m∠̇C = 13º.

Problem 9 :

Find m∠̇A

missing-sides-angles-using-law-of-sines-q9

Solution :

CB = a = 11, AC = b = 33, AB = c = 25

∠̇B = 129º 

asin A = bsin B = csin C 11sin A = 33sin 129° = 25sin C 11sin A = 33sin 129° 11sin A = 330.77711sin A =42.47 sin A = 1142.47sin A = 0.26A = sin -1 (0.26)A = 15°

Hence, the required angle is m∠̇A = 15º.

Problem 10 :

Find m∠̇C

missing-sides-angles-using-law-of-sines-q10

Solution :

CB = a = 22, AC = b = 9, AB = c = 19

∠̇A = 97º 

asin A = bsin B = csin C 22sin 97° = 9sin B = 19sin C 22sin 97° = 19sin C 220.99 = 19sin C22.22 = 19sin C sin C = 1922.22sin C = 0.855C = sin -1 (0.855)C = 58.76C = 59°

Hence, the required angle is m∠̇C = 59º.

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