Download Lesson 2 - sunmath

Lesson 2.7 – Trigonometry in the Real World
Problem #1:
While walking to school you pass a barn with a silo. Looking up to the top of the
silo you estimate the angle of elevation to the top of the silo to be about 14°. You
continue walking and find that you were around 40 m from the silo. Using this
information and your knowledge of trigonometric ratios calculate the height of the
silo.
A
14°
B
40 m
C
Solution:
While walking to school you pass a barn with a silo. Looking up to the top of the
silo you estimate the angle of elevation to the top of the silo to be about 14°. You
continue walking and find that you were around
40 m from the silo. Using this information and your knowledge of trigonometric
ratios calculate the height of the silo.
A
14°
B
40 m
C
Side b represents the height of the silo
opposite
tan B 
adjacent
b
tan B 
a
b
tan 14o 
40
o
40(tan 14 )  b
40(0.2493...)  b
9.97312...  b
10.0  b
The silo is approximately 10.0m tall
Problem #2:
David wants to go to Toronto from
Edmonton, but he took the wrong road and
ended up in Chicago instead. Upon realizing
his directional mistake, David drove from
Chicago to Toronto. If the angle at Toronto is
45°, the angle at Chicago is 95°, and the
distance from Edmonton to Toronto is 2000
km, how much further did David drive than
necessary?
E
2000 km
T
45O
95O
C
Solution:
The
 E can be found using the sum of the angles in a triangle:
E  180o  95o  45o
E  40o
Now use Sine Law to find the remaining two sides:
e
t
c


sin E sin T sin C
e
t
2000


sin 40 sin 45 sin 95
Break this into two separate equations and solve for e and t.
e
2000

o
sin 40
sin 95o
2000(sin 40o )
e
sin 95o
2000(0.64278...)
e
0.996194...
1285.57521...
e
0.996194...
e  1290.4859...
t
2000

o
sin 45
sin 95o
2000(sin 45o )
t
sin 95o
2000(0.70710...)
t
0.996194...
1414.21356...
t
0.996194...
t  1419.6156...
e  1290.5km
t  1419.6km
Total distance travelled = 1290.5 + 1419.6
= 2710.1 km
Extra distance travelled = 2710.1 – 2000
= 710.1 km
 David drove 710.1 km more than was necessary
Problem #3:
Jill and her friends built an outdoor hockey rink. Their hockey goal line is 5 feet wide.
Jill shoots a puck from a point where the puck is 5 yards from one goal post and 6 yards
from the other goal post. Within what angle must Jill make her shot to hit the net?
Keep in mind that there are 3 feet in a yard.
5 yards
5 feet
6 yards
Solution:
Jill and her friends built an outdoor hockey rink. Their hockey goal line is 5 feet wide. Jill shoots
a puck from a point where the puck is 5 yards from one goal post and 6 yards from the other goal
post. Within what angle must Jill make her shot to hit the net?
B
5 yards = 15 feet
5 feet
C
6 yards = 18 feet
A
If we make the position where Jill is standing A and the goalposts B and C then in this case we
can use the Cosine Law to solve for the angle.
b2  c2  a 2
2bc
2
18  152  52
cos A 
2(18)(15)
324  225  25
cos A 
540
cos A  0.9703703...
cos A 
A  cos 1 0.9703703...
A  13.98...o
A  14o
 Jill must shoot within an angle of about 14° to hit the net.