Transcript Trigonometry Sine Area Rule

Trigonometry
Sine Area Rule
B
1
Area = acsin B
2
a
C
c
1
Area = bcsin A
2
b
A
By Mr Porter
1
Area = absinC
2
Definition: The area of any triangle, ABC, can be found by using the formula:
1
Area = absinC
2
B
b
c
A
a
C
The formula is cyclic:
1
Area = absinC
2
1
Area = bcsin A
2
1
Area = casin B
2
Example 1: Calculate the area of triangle
ABC, correct to 3 sig. fig.
B
Label the sides of the
triangle a, b and c..
b
55 m
c
42°15’
A
62 m
a
Example 2: Calculate the area of triangle
ABC, correct to 1 dec. pl..
r 12.5 cm
Q
P
108°40’
Label the side of
13.8 cm
the triangle p, q
p
and r.
q
Write the the formula for this triangle.
C
Area =
1
pr sinQ
2
Write the the formula for this triangle.
1
Area = absinC
2
Substitute the values for a, b and angle C.
Area =
1
´ 62 ´ 55 ´ sin(42°15')
2
Evaluate and round to 3 sig. fig.
Area = 1146.385407
Area = 1150m 2
Substitute the values for p, r and angle Q.
Area =
1
´13.8 ´12.5 ´ sin(108°40')
2
Evaluate and round to 1 dec. pl.
Area = 81.71296042
Area = 81.7cm 2
R
Example 3: Calculate the area of
quadrilateral, correct to 3 sig. fig.
c2 = a2 + b2
C
B
35 m
15 m
A
20 m
Side c is missing! (Hypotenuse of ∆ABD)  Pythagoras’ Thm.
Note: To find the
area of the
quadrilateral, you
need to calculate the
area of the 2
triangles.
D
Label Quadrilateral (Triangles)
Area of triangle ABD,
AABD = 1/2 x B x H
AABD = ½ x 20 x 15
AABD = 150 m2
Area of ∆BDC, sine area formula
1
Area = bcsin D
2
c2 = 202 + 152
c2 = 625
c =√625
c = 25
Side BD is 25 m.
Area of ∆BDC, sine area formula
1
Area = bcsin D
2
1
Area = ´ 25 ´ 35 ´ sin(55°30')
2
Area = 360.555 m 2
Hence,
the total area = 150 + 360.555
= 510.555
= 511 m2
Example 4: Calculate the area of regular
hexagon of radius 12 m, correct to 3 sig. fig.
Draw a diagram with radial diagonals.
θ = 60°
Example 5: Calculate the area of regular
pentagon of radius 15 cm, correct to 3 sig. fig.
Draw a diagram with all radii from the centre drawn.
Angle at the centre for a
regular polygon is
θ = 360÷ number of sides.
Angle at the centre for a
regular polygon is
θ = 360÷ number of sides.
12 m
θ = 72°
Area is 6 times the area of 1 triangle.
60°
Area is 5 times the area of 1 triangle.
12 m
Area of ∆ABC, sine area formula
1
Area = absinC
2
Substitute the values for a, b and angle C.
Area of ∆ABC, sine area formula
1
Area = absinC
2
1
´12 ´12 ´ sin 60°
2
1
Total Area = 6 ´ ´12 ´12 ´ sin 60°
2
Total Area = 748 m 2
1
´15 ´15 ´ sin 72°
2
1
Total Area = 5 ´ ´15 ´15 ´ sin 72°
2
Total Area = 535 m 2
Area =
Substitute the values for a, b and angle C.
Area =
Example 6: The results of a radial survey are
shown in the diagram (all measurements in
metres). Calculate the total area of ∆XYZ
(nearest m2 ).
North
(0°T)
Now apply the sine area rule 3 times.
Area of ∆XOZ: Area =
Area =
X (045°T, 68m)
1
xz sinq
2
1
´ 92 ´ 68 ´ sin130°
2
Area = 2396.187 m 2
Likewise, areas for ∆XOY and ∆YOZ
Z
(275°T, 92 m)
O
Area of ∆XOY= 3258.854 m2
Y (155°T, 102m)
Need to calculate the area of all 3 triangles.
Must calculate the angles around ‘O’.
From their bearing: Angle XOY = 155 – 45 = 110°
Angle YOZ = 275 – 155 = 120°
Now, Angle XOZ = (360 – 275) + 45 = 130°
OR, Angle XOZ = 360 – (110 + 120) = 130°
Area of ∆YOZ= 4063.391 m2
Hence,
total area = 2396.187 + 3258.854 + 4063.391
= 9718.432 m2
= 9718 m2