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secondary 4 | E Maths
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hey teachers i really struggle with the questions . please help
Date Posted: 3 years ago
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Qn 11a
We'll be using the cosine rule a^2 = b^2 + c^2 - 2bcCosA
Therefore
9^2 = 5^2 + 6^2 - 2(5)(6)CosBAC
60 Cos BAC = 61 - 81
Cos BAC = - 1/3
BAC = Cos-1 (-1/3)
BAC = 109.47° (2 d.p)
11b
Angle POR acute = 360 - 250 = 110°
Angle PQR = 110° / 2 = 55°
Look up your syllabus on angles in a circle
We'll be using the cosine rule a^2 = b^2 + c^2 - 2bcCosA
Therefore
9^2 = 5^2 + 6^2 - 2(5)(6)CosBAC
60 Cos BAC = 61 - 81
Cos BAC = - 1/3
BAC = Cos-1 (-1/3)
BAC = 109.47° (2 d.p)
11b
Angle POR acute = 360 - 250 = 110°
Angle PQR = 110° / 2 = 55°
Look up your syllabus on angles in a circle
11bii
pi rad = 180°
1° = (pi / 180) rad
250° = (pi x 250 / 180) rad
250° = 4.363 rad (3 d.p)
biii
Area of circle = pi x r^2 = 64pi
Area of major sector POR = 250 / 360 x 64pi
= 139.63 cm² (2 d.p)
Alternatively in rad
Area of circle = 1/2 x r^2 x Angle in rad
Area of major sector POR = 0.5 x r^2 x 4.363
= 139.62 cm² (2 d.p)
pi rad = 180°
1° = (pi / 180) rad
250° = (pi x 250 / 180) rad
250° = 4.363 rad (3 d.p)
biii
Area of circle = pi x r^2 = 64pi
Area of major sector POR = 250 / 360 x 64pi
= 139.63 cm² (2 d.p)
Alternatively in rad
Area of circle = 1/2 x r^2 x Angle in rad
Area of major sector POR = 0.5 x r^2 x 4.363
= 139.62 cm² (2 d.p)
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