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secondary 4 | E Maths
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Date Posted: 1 year ago
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Using cosine rule,
AC² = AB²+BC²–2(AB)(BC)cosABC
430² = 367²+266²–2(367)(266)cosABC
cosABC = 0.10523 (5s.f.)
ABC = 84.0° (1d.p) (Shown)
Let N1 be north of A, N2 be north of B, N3 be north of C
N2BA = 180°–62° = 118° (angle between // lines)
N2BC = 360°–118°–84.0° = 158° (angle at a point)
N3CB = 180°–158° = 22° (angle between // lines)
Bearing of B from C = 360°–22° = 338° (angle at a point) (Shown)
a)
Using sine rule,
AC/sinABC = BC/sinBAC
430/sin84.0° = 266/sinBAC
sinBAC = 0.61522 (5s.f.)
BAC = 37.968° (3d.p.) = 38.0° (1d.p.)
b)
Area of triangle ABC = ½(AB)(BC)sinABC = ½(367)(266)sin84.0° = 48544m² (5s.f.) = 48500m² (3s.f.)
ci)
tanACD = AD/AC
tan12° = AD/430
AD = 91.399m² (5s.f.) = 91.4m²
cii)
Let the point on BC that has the shortest distance from A be E m
ACB = 180°–37.968°–84° = 58.032° (3d.p.) (angles in a triangle)
sinACB = AE/AC
sin58.032 = AE/430
AE = 364.79m (5s.f.) = 365m (3s.f.)
Shortest distance from A to BC = 365m
ciii)
tanAED = AD/AE
tanAED = 91.399/364.79
AED = 14.066° (3d.p) = 14.1° (1d.p.)
Greatest angle of elevation = 14.1°
AC² = AB²+BC²–2(AB)(BC)cosABC
430² = 367²+266²–2(367)(266)cosABC
cosABC = 0.10523 (5s.f.)
ABC = 84.0° (1d.p) (Shown)
Let N1 be north of A, N2 be north of B, N3 be north of C
N2BA = 180°–62° = 118° (angle between // lines)
N2BC = 360°–118°–84.0° = 158° (angle at a point)
N3CB = 180°–158° = 22° (angle between // lines)
Bearing of B from C = 360°–22° = 338° (angle at a point) (Shown)
a)
Using sine rule,
AC/sinABC = BC/sinBAC
430/sin84.0° = 266/sinBAC
sinBAC = 0.61522 (5s.f.)
BAC = 37.968° (3d.p.) = 38.0° (1d.p.)
b)
Area of triangle ABC = ½(AB)(BC)sinABC = ½(367)(266)sin84.0° = 48544m² (5s.f.) = 48500m² (3s.f.)
ci)
tanACD = AD/AC
tan12° = AD/430
AD = 91.399m² (5s.f.) = 91.4m²
cii)
Let the point on BC that has the shortest distance from A be E m
ACB = 180°–37.968°–84° = 58.032° (3d.p.) (angles in a triangle)
sinACB = AE/AC
sin58.032 = AE/430
AE = 364.79m (5s.f.) = 365m (3s.f.)
Shortest distance from A to BC = 365m
ciii)
tanAED = AD/AE
tanAED = 91.399/364.79
AED = 14.066° (3d.p) = 14.1° (1d.p.)
Greatest angle of elevation = 14.1°
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