Victoria University MATH 132 Assignment 6: Trigonometry and Triangles

This document presents a comprehensive solution to Assignment 6 for MATH 132, focusing on fundamental trigonometric concepts and their applications. The solution begins by calculating trigonometric ratios (sine, cosine, and tangent) using Pythagorean triples and the Law of Sines, demonstrating the relationships between angles and sides in right-angled triangles. It then applies the Pythagorean theorem to determine unknown side lengths in various triangles. The assignment also explores geometric constructions, showing how a right-angled triangle can be divided into two isosceles triangles. Furthermore, the solution utilizes the unit circle to calculate trigonometric ratios for different angles, including sine, cosine, and tangent values. It continues by calculating the area of a triangle using the formula involving sine and finds the length of a dotted line, applying Pythagoras' theorem. The solution concludes by testing triangle inequalities to determine if given side lengths can form triangles and calculating the area of triangles using Heron's formula. The document contains figures and diagrams to illustrate each concept. Finally, the solution provides tutorial exercises with similar questions and solutions.

( 1,1 , √ 2 )and ( 1 , √ 3 , 2 )are Pythagorean triples, so triangles with these sides form right angled
triangles as shown in figure above.
Using the SINE law for triangle ( 1,1 , √ 2 ):
1
sin ( c ) = 1
sin ( b ) = √ 2
sin ( 90 )
Implies,
∠ b=∠c
but,
In a triangle abc : ∠a+∠ b+∠c=180 °
therefore,
90 ° +∠b+∠b=180 ° ⇒ ∠b=∠ c=45 °
Using the SINE law for triangle ( 1 , √ 3 , 2 ):
1
sin ( r ) = 2
sin ( 90 ) = √ 3
sin ( p )
therefore,
sin ( r ) =1 ·sin ( 90 )
2 ⇒ sin ( r ) =1
2 ⇒ r=sin−1
( 1
2 )=30 °
∠ p+∠q+ ∠r=180 °
⇒ 90 °+∠ q +30 °=180 ° ⇒ ∠ q=60 °
a) sin ( 60 ° )

From triangle pqr:
sin ( 60 ° ) =sin ( q ) =side opposite ¿ q ¿
hypotenuse = pr
qr = √ 3
2
b) cos ( 45 ° )
From triangle abc:
cos ( 45 ° ) =cos ( c ) =side adjecent ¿ c ¿
hypotenuse = ac
bc = 1
√ 2
c) tan ( 30 ° )
From triangle pqr:
tan ( 30 ° )=tan ( r )=side opposite ¿ r ¿
side adjecent ¿ r ¿= 1
√3
The Pythagoras rule for right angled triangles states that:
Square of hypotenuse is equal to the sum of squares of remaining two sides.
Hypotenuse is the side opposite to the right angle, which is ‘a’ for first triangle and ‘9’ for the second
triangle.
Using Pythagoras rule for the first triangle:
a2=52 +32
⇒ a= √ 52 +32= √ 34 ≅ 5.83
Using Pythagoras rule for the second triangle:
92 =z2+82 ⇒ z2=92−82 =17⇒ z= √ 17 ≅ 4.12

Consider any right angle triangle abc as shown in the figure above.
Circumcentre of a right angle triangle lies at the midpoint of hypotenuse (=ac).
Point d is the midpoint of hypotenuse ac. As d is the circumcentre, points a, b and c lie on a
circumcircle with radius = da or db or dc. Thus, da = db = dc which is indicated by the labelling in
figure above.
This means that triangle adb is a isosceles triangle with ad = bd, and triangle cdb is also an isosceles
triangle with bd = cd. Thus, with this simple construction, we end up with 2 isosceles triangles from a
right angle triangle.
Equal lengths and equal angles are marked in the figure below:

Trigonometric ratios on a unit circle (radius =1, centred at origin) are shown in figure below.
Form the figure we deduce that for an angle θ, the x- coordinate of the radius at that angle is cos θ,
y- coordinate is sin θ and tanθ is the length of tangent from the point upto the x- axis.
Graph of unit circle showing various angles and coordinates for each of the angles (point of
intersection of radius at the angle and the circle), is shown in the figure below.
a) sin ( 210 ° )

sin ( 210 ° )=−1
2
b) cos ( 135 ° )
cos ( 135 ° ) =− √ 2
2
c) tan ( 330 ° )
tan ( 330 ° )=−1
√ 3

a)
Area of triangle:
A=1
2 ab· sin ( C )
a=4 , b= √2 , ∠C=45 °
therefore,
A=1
2 ( 4 ) · ( √2 ) · sin ( 45 )=2 · √2 · 1
√ 2=2
b)
Length of dotted line (p):
p is perpendicular to side a, therefore sides p,q and b for a right angle triangle. Side b is the
hypotenuse.
therefore,
sin ( 45 ° ) = p
b ⇒ p=b· sin ( 45 ° )= √2 · 1
√2 =1
Also,
cos ( 45 ° ) = q
b ⇒ q=b· cos ( 45 ° ) = √ 2 · 1
√ 2 =1

For the right angle triangle with sides as shown in figure above, by Pythagoras’ rule:
c2=32+12 ⇒c= √ 32 +12= √ 10
Triangle inequality test: the length of any side of a triangle is greater than the absolute difference of
the lengths of the other two sides.
a) 3, 4, 5.
3 > |4-5| is true
4>|3-5|, true
5>|3-4|, true
So, these lengths represent sides of a triangle.
Area of triangle:
A= √ s ( s−a ) ( s−b ) ( s−c )
here, a=3, b=4, c=5 and
s= a+ b+c
2 = 3+4 +5
2 =6
therefore,
A= √6 ( 6−3 ) ( 6−4 ) ( 6−5 )= √6 ·3 · 2· 1= √36=6
b) 4, 5, 7.
4 > |5-7|, true
5 > |4-7|, true
7 > |4-5|, true

So, these lengths represent sides of a triangle.
Area of triangle:
a = 4, b = 5, c = 7 and
s= a+ b+c
2 = 4 +5+7
2 =8
therefore,
A= √8 ( 8−4 ) ( 8−5 ) ( 8−7 )= √ 8 · 4 · 3· 1= √9 6
c) 5, 6, 12.
5 > |6-12|, is false. So the test fails and these sides don’t represent a triangle.
Tutorial Exercise
( 1,1 , √ 2 )and ( 1 , √ 3 , 2 )are Pythagorian triples, so triangles with these sides form right angled
triangles as shown in figure above.
Using the SINE law for triangle ( 1,1 , √ 2 ):
1
sin ( c ) = 1
sin ( b ) = √ 2
sin ( 90 )
Implies,
∠ b=∠c
but,
In a triangle abc : ∠a+∠ b+∠c=180 °

therefore,
90 ° +∠b+∠b=180 ° ⇒ ∠b=∠ c=45 °
Using the SINE law for triangle ( 1 , √ 3 , 2 ):
1
sin ( r ) = 2
sin ( 90 ) = √ 3
sin ( p )
therefore,
sin ( r ) =1 ·sin ( 90 )
2 ⇒ sin ( r ) =1
2 ⇒ r=sin−1
( 1
2 )=30 °
∠ p+∠q+ ∠r=180 °
⇒ 90 °+∠ q +30 °=180 ° ⇒ ∠ q=60 °
a) sin ( 30 ° )
From triangle pqr:
sin ( 30 ° )=sin ( r ) =side opposite to angle r
hypotenuse = pq
qr =1
2
b) tan ( 45 ° )
From triangle abc:
tan ( 45 ° ) =tan ( c ) =side opposite to angle c
side adj a cent to angle c = ab
ac = 1
1=1
a)
Property of isosceles triangles : the angle bisector of the vertex angle is also the perpendicular
bisector of the base.

Isosceles triangle is shown in figure with two sides equal (=a). Side c is the base of the triangle and
angle C is the vertex angle. Therefore, by property of isosceles triangle, angle bisector of C (red line)
is perpendicular to base c and divides it into equal lengths.
b) Consider any one of the right angle triangle (shown in figure below)
For the right angle triangle, we have:
cos ( A ) =side adjacent to A
Hypotenuse =
c
2
a ⇒ cos ( A ) = c
2 a ⇒ c=2 a cos ( A )
Trigonometric ratios on a unit circle (radius =1, centred at origin) are shown in figure below.
Form the figure we deduce that for an angle θ, the x- coordinate of the radius at that angle is cos θ,
y- coordinate is sin θ and tanθ is the length of tangent from the point upto the x- axis.
Graph of unit circle showing various angles and coordinates for each of the angles (point of
intersection of radius at the angle and the circle), is shown in the figure below.

a) cos ( 15 0 ° )
cos ( 150 ° )=− √3
2
b) tan ( 225 ° )
tan ( 225 ° ) =1

a)
Area of triangle:
A=1
2 ab· sin ( C )
a=7 , b=2 √3 , ∠C=30°
therefore,
A=1
2 (7 ) · ( 2 √3 ) · sin ( 30 ) =7 · √3 · 1
2 =7 √3
2
b)
Length of dotted line (p):
p is perpendicular to side a, therefore sides p,q and b for a right angle triangle. Side b is the
hypotenuse.
therefore,
sin ( 30 ° )= p
b ⇒ p=b·sin ( 30 ° )=2 √3 · 1
2 = √3
Also,
cos ( 30 ° )= q
b ⇒q=b· cos ( 30 ° )=2 √3 · √3
2 =3