Download PTG 0016 Note 1.3

TOPIC 1.3 – APPLICATIONS
OF TRIGONOMETRIC
FUNCTIONS
1.3.1: The Law of Sines
1.3.2: The Law of Cosines
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1.3.1: The Law of Sines
Definition:
A triangle is called oblique, if none of the angles of the
triangle is a right angle.
To solve an oblique triangle means to find the lengths of its
sides and the measurements of it angles. Four possibilities
have to be considered:
1. One side and two angles are known. (SAA)
2. Two sides and the angle opposite one of them are known.
(SSA)
3. Two sides and the included angle are known. (SAS)
4. Three sides are known. (SSS)
The first two cases require the Law of Sines and the last two
requires the Law of Cosines.
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The law of sines
If A, B and C are the measures of the angles of a triangle a,
b, and c are the length of the sides opposite these angles,
then
a
b
c
=
=
sin A sin B sin C
The ratio of the length of the side of any triangle to the sine of
the angle opposite that side is the same for all three sides of the
triangle
Solving an SAA using law sines
Example:
1.Solve the triangle with A= 64º, C = 82º and c = 14cm
2.Solve triangle ABC if A = 40º, C = 22.5° and b = 12
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The ambiguous case SSA
When solving triangle where two sides and the angle opposite one of
them are known, the result will lead to one triangle, two triangles or no
triangle at all. ( a, b are sides of the triangle and h is the height).
If a < b sin α = h, then
No Triangle
clearly side a is not
sufficiently long to form a
triangle
If a = b sin α = h, then side a One Right
Triangle
is just long enough to
form a right triangle
If a < b and h = b sin α < a,
Two
Triangles
then two distinct triangles
can be formed.
If a ≥ b, then only one
triangle can be formed
One
Triangle.
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Solving an SSA using law sines (One solution)
Example:
Solve triangle ABC if A = 123º a = 47 and c = 23
Solving an SSA using law sines (No solution)
Example:
Solve triangle ABC if A = 50º a = 10 and b = 20
Solving an SSA using law sines (Two solutions)
Example:
Solve triangle ABC if A = 35º a = 12 and b = 16
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Area of an oblique triangle
1
1
1
Area,
A = bc sin A = ab sin C = ac sin B
2
2
2
Example:
Find the area of a triangle having two sides of lengths 8 meters and
12 meters and included angle of 135°
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10.2: The Law of Cosine
The law of cosines:
If A, B and C are the measures of the angles of a
triangle and a, b and c are the length of the sides
opposite these angles, then
a 2 = b 2 + c 2 − 2bc cos A
b 2 = a 2 + c 2 − 2 ac cos B
c 2 = a 2 + b 2 − 2 ab cos C
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Solving an SAS triangle
1. Use the law of cosines to find the side opposite the given angle
2. Use the law of sines to find the angle opposite the shorter of
the two given sides. This angle is always acute.
3. Find the third angle by subtracting the measure of the given
angle and the angle found in step 2 from 180º
Example:
Solve the triangle with A = 120º, b = 7 and c = 8
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Solving an SSS triangle
1.Use the law of cosines to find the angle opposite the longest
side
2.Use the law of sine to find either of the two remaining acute
angles
3.Find the third angle by subtracting the measures of the angles
found in step 1 and 2 from 180°
Example:
Solve triangle ABC if a = 8, b = 10 and c = 5. Round to the nearest tenth of
a degree.
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Heron’s formula for the area of triangle:
The area of the triangle with sides a, b and c is
Area =
where
s=
s ( s − a )( s − b)( s − c)
1
( a + b + c)
2
Example:
Find the area of the triangle with a = 6m, b = 16m and c = 18m.
Round to the nearest square meter.
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Summary
Law of Sines
a
b
c
=
=
sin A sin B sin C
Area of an oblique triangle
1
1
1
Area,
A = bc sin A = ab sin C = ac sin B
2
The law of cosines:
a 2 = b 2 + c 2 − 2bc cos A
b 2 = a 2 + c 2 − 2 ac cos B
2
2
Heron’s formula for the area of triangle:
Area =
s ( s − a )( s − b)( s − c)
c 2 = a 2 + b 2 − 2 ab cos C
where
s=
1
( a + b + c)
2
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