Solving Triangles

Solutions of Triangles

Institute of Mathematics, University of the Philippines Diliman

Meeting 27

(IMath, UPD) Solutions of Triangles Meeting 27 1 / 23

Outline

1 Triangles

2 Solving Right Triangles

3 Solving Oblique Triangles

4 Applications


Triangles

Notation for triangles

1 A, B, C: vertices

2 a, b, c: lengths of the respective opposite sides

3 ↵, �, �: measures of the respective interior angles


Triangles

1 A, B, C: vertices

2 a, b, c: lengths of the respective opposite sides

3 ↵, �, �: measures of the respective interior angles

Recall:

↵+ � + � = 180

�

Triangle Inequality The sum of the lengths of two sides of a triangle isgreater than the length of the third side.

Hinge Theorem In a 4, the side opposite the larger angle is thelonger side.

Pythagorean Theorem If 4ABC is a right triangle with hypotenuseof length c, then a2 + b2 = c2.


Definition

A triangle that is not a right triangle is called an oblique triangle.

GOAL:

Solving a triangle means solving for the length of its sides and the measureof its interior angles.


Solving Right Triangles

✓: acute angle in a right triangle

Place the 4 on the Cartesian plane such that ✓ is in standard position

✓

adjacent

opposite

h

y

p

o

t

e

n

u

s

e

x = adjacent, y = opposite, r = hypotenuse

sin ✓ =

opposite

hypotenuse

, cos ✓ =

adjacent

hypotenuse

, tan ✓ =

opposite

adjacent



Theorem

For an acute angle ✓ in a right triangle, we have

sin ✓ =

opposite

hypotenuse

cos ✓ =

adjacent

hypotenuse

tan ✓ =

opposite

adjacent



Example: Solve the triangle with b = 12, � = 15

�,and � = 90

�.

Solution:↵ = 180

� � (90

�+ 15

�) = 75

�

sin 15

�=

c

12=) c = 12 sin 15

�= 12 sin(45

� � 30

�)

= 12 (sin 45

�cos 30

� � cos 45

�sin 30

�)

= 12

⇣p22 ·

p32 �

p22 · 1

2

⌘

= 12

⇣p6�

p2

4

⌘= 3

p6� 3

p2

cos 15

�=

a

12 =) a = 12 cos 15

�= cos(45

� � 30

�)

=) a = 12 (cos 45

�cos 30

�+ sin 45

�sin 30

�)

=) a = 12

⇣p6+

p2

4

⌘= 3

p6 + 3

p2


Sine Law

Drop a perpendicular segment from C to AB.Then, sin↵ =

h

b

and sin� =

h

a

. Hence,

h = b sin↵ = a sin�sin↵

a=

sin�

b

Theorem (Sine Law)

For a triangle with interior angles ↵,�, � opposite sides with lengths a, b, c

respectively:

sin↵

a=

sin�

b=

sin �

c


Example: Solve the triangle with c = 8, ↵ = 45

�, and � = 60

�.

Solution:� = 180

� � (60

�+ 45

�) = 75

�

sin 75

�

8

=

sin 60

�

b) b =

8 sin 60

�

sin 75

� =

8

⇣p32

⌘

p6+

p2

4

= 12

p2� 4

p6

sin 75

�

8

=

sin 45

�

a) a =

8 sin 45

�

sin 75

� =

8

⇣p22

⌘

p6+

p2

4

= 8

p3� 8


Cosine Law

Set 4ABC on the Cartesian plane as shown below:

Let (x, y) be the coordinates of A. Then

cos � =

x

b

and sin � =

y

b

Therefore, (x, y) = (b cos �, b sin �).


Cosine Law

By the distance formula,

c2 = (b cos � � a)2 + (b sin � � 0)

2

c2 = b2 cos2 � � 2ab cos � + a2 + b2 sin2 �c2 = a2 + b2(sin2 � + cos

2 �)� 2ab cos �c2 = a2 + b2 � 2ab cos �


Cosine Law

By changing the side of the triangle fixed on the positive x-axis, we get:

Theorem (Cosine Law)

a2 = b2 + c2 � 2bc cos↵

b2 = a2 + c2 � 2ac cos�

c2 = a2 + b2 � 2ab cos �

Note: When one of the angles, say �, is 90�, the formula becomesc2 = a2 + b2.


Example: Solve the triangle with a = 10, b = 10, and c = 10

p3.

Solution:(10

p3)

2= 10

2+ 10

2 � 2(10)(10)(cos �) ) cos � = �12

) � = Arccos

�� 1

2

�= 120

�

Since 4ABC is isosceles, ↵ = �.

) ↵ = � =

180��120�

2 = 30

�


Solve all triangles ABC satisfying a = 18, c = 8, � = 30

�, if any.

sin↵

18

=

sin 30

�

8

sin↵ =

18

�12

�

8

=

9

8

> 1

NO SOLUTION!


Solve all triangles ABC satisfying b = 5

p2, c = 5

p3, � = 60

�, if any.

sin�

5

p2

=

sin 60

�

5

p3

) sin� =

5

p2

�p32

�

5

p3

=

p2

2

) � = 45

� or 135�. But, if � = 135

�, then � + � > 180

�.

� = 45

�,↵ = 180

� � (45

�+ 60

�) = 105

�

a

sin 105

� =

5

p3

sin 60

� ) a =

5

p3

�p6+

p2

4

�p32

=

5

�p6 +

p2

�

2


Applications

Definition

E: observer’s eye, O: point being observed

line of sight – line joining E and O

angle formed by the horizontal and the line of sight:angle of elevation, if O is above the horizontal through Eangle of depression, if O is below the horizontal through E


Example: Roni, standing 6 ft away from a lamp post, observed that theangle of elevation to the top of the lamp post is 30�, while the angle ofdepression to the bottom of the lamp post is 45�. How high is the lamppost?Solution:

tan 30

�=

a

6

) a = 6 tan 30

�= 6

⇣p33

⌘= 2

p3

tan 45

�=

b

6) b = 6 tan 45

�= 6(1) = 6

height = a+ b = 6 + 2

p3


Example: Lana the Ant observed that the top of a jar of sugar has anangle of elevation of 30�. After walking 1 ft towards the jar, the new angleof elevation is 60�. What is the height of the jar?Solution:

tan 60

�=

h

a

) a =

h

tan 60� =

hp3

tan 30

�=

h

a+1 ) h =

p33 (a+ 1)

) h =

p33 (

hp3+ 1) =

h

3 +

p33

) 2h3 =

p33

h =

p32 ft


Definition

O, P : points

bearing or direction of P from O: measure of the acute angle whichOP makes with the north–south line

course (navigation): measure of the angle formed by the path of aplane or ship with the line due north, measured clockwise

First figure:bearing of P from O: N75�EThe path to P from O has course 75

�.


Definition

O, P : points

bearing or direction of P from O: measure of the acute angle whichOP makes with the north–south line

course (navigation): measure of the angle formed by the path of aplane or ship with the line due north, measured clockwise

Second figure:bearing of P from O: S60�WThe path to P from O has course 240

�.


Example: A ship traveled 10 km N30�E. Then, it changed its course toS15�E and traveled a distance of 20

p2 km. How far is the ship now from

its origin?

Solution:

Let c be the distance of the shipfrom its origin.

c2 = 10

2+ (20

p2)

2

� 2(10)(20

p2) cos 45

�

c2 = 100 + 800

� 2(10)(20

p2)

⇣p22

⌘

c2 = 900� 400

c2 = 500

c = 10

p5 km


EXERCISES:1 Solve all triangles ABC (if any) satisfying:

1 a = 3, b = 4, c = 7

2 a =

p6, b = 2,� = 45

�

2 A tree stands vertically on top of a hill whose angle of inclination is30

�. The top and base of the tree are 4

p3 and 4 meters, respectively,

from a point P at the foot of the hill. What is the angle subtendedby the tree at point P?

3 At a distance of 6 feet from the base of the Oblation, the angle ofelevation of the head and feet of the statue (on the pedestal) are 60

�

and 30

�, respectively. How high is the statue?

4 Two airplanes leave an airport at the same time. Plane A flies on acourse of 35� at 150 kph and Plane B moves on a course of 125� at200 kph. Find after 2 hours, (a) the distance between the two planes,(b) the bearing from Plane A to Plane B and (c) the bearing fromPlane B to Plane A.


Solving Triangles

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