In Problems 1 and 2, use the cosine and sine
rules to solve the triangles PQR and find their
areas.
1. q = 12 cm,r = 16 cm, P = 54◦
p = 13.2 cm, Q = 47◦21
,
R = 78◦39
, area = 77.7 cm2
2. q = 3.25 m,r = 4.42 m, P = 105◦
p = 6.127 m, Q = 30◦50
,
R = 44◦10
, area = 6.938 m2
Answers
Step-by-step explanation:
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Problem 1
q = 12, r = 16, P = 54°
Cosine rule:
p² = q² + r² - 2qr cos P
=> p = √ (12² + 16² - (2)(12)(16) cos 54°) ≈ 13.2
Sine rule:
sin Q / q = sin P / p
=> sin Q = q sin P / p
≈ 12 sin 54° / 13.2
=> Q ≈ arcsin ( 12 sin 54° / 13.2 )
≈ 47.34° (This is with taking care to avoid round-off error. It is very difficult to guess exactly what errors lead to 47.21°.)
Sine rule again:
sin R / r = sin P / p
=> sin R = r sin P / p
=> R = arcsin ( r sin P / p )
≈ arcsin ( 16 sin 54° / 13.2 )
≈ 78.66° (again with care to avoid round-off error)
Area
= (1/2) q r sin P
= 0.5 × 12 × 16 × sin 54°
≈ 77.7
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Problem 2
q = 3.25, r = 4.42, P = 105°
Cosine rule:
p² = q² + r² - 2qr cos P
=> p = √ (3.25² + 4.42² - (2)(3.25)(4.42) cos 105°) ≈ 6.127
Sine rule:
sin Q / q = sin P / p
=> sin Q = q sin P / p
≈ 3.25 sin 105° / 6.127
=> Q ≈ arcsin ( 3.25 sin 105° / 6.127 )
≈ 30.82° (taking care to avoid round-off error)
Sine rule again:
sin R / r = sin P / p
=> sin R = r sin P / p
=> R = arcsin ( r sin P / p )
≈ arcsin ( 4.42 sin 105° / 6.127 )
≈ 44.18° (again with care to avoid round-off error)
Area
= (1/2) q r sin P
= 0.5 × 3.25 × 4.42 × sin 105°
≈ 6.938
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