Aim: How do we apply the trig ratios for indirect measurements?
Objectives: to apply the trig ratios to calculate unknown length when given a right triangle
Lesson Development: We established the special relationship between the side lengths of the 45-45-90
triangle and the 30-60-90 triangle. Is there something for non 45-45-90 and 30-60-90 right triangles?
It turns out there is. There are called trig ratios. Let’s go over some very important terminologies first.
C is a right angle. When we are from the perspective of A , we say BC is the opposite side and
AC is the adjacent side and AB , which faces the right angle, is the hypotenuse.
From the perspective of B , then BC is the adjacent side while AC is the opposite side. AB remains
the hypotenuse.
Note: The adjacent side cannot be the hypotenuse.
Tangent Ratio: tan A 
Sine Ratio: sin A 
length of side opposite A
a

length of side adjacent to A b
length of side opposite A a

length of hypotenuse
c
Cosine Ratio: cos A 
length of side adjacent A b

length of hypotenuse
c
b
a
b
All three trig ratios are using A . If we use B , sin B  , cos B  , tan B 
c
c
a
Note: trig ratios only work with right triangles as the Pythagorean Theorem only applies to right
triangle.
One of the legs = 1
sin 30 
1
2
3
2
1
3
tan 30 

3
3
cos 30 
Short leg = 1
1
2

2
2
1
cos 45 
2
tan 45  1
sin 45 
3
2
1
cos 60 
2
sin 60 
tan 60  3
Notice we can determine tan 30 by knowing the values of sin 30 and cos30 ? Definition:
sin 
tan  
cos 
A and B can represent the measure of the interior angles other than 30, 45 and 60 degrees. Now let’s
apply the trig ratios.
EX1: Solve for x and y to the nearest hundredth
y
y
sin 36 
cos 54 
7
7
y  4.11
y  4.11
or
x
x
sin54=
cos36=
7
7
x  5.66
x  5.66
EX2: Solve for a and b to the nearest tenth
sin 22 
b
30
b  11.2
cos 22 
a
30
a  27.8
EX3: A surveyor needs to determine the distance
across the pond shown in the accompanying
diagram. She determines that the distance from
her position to point P on the south shore of the
pond is 175 meters and the angle from her
position to point X on the north shore is 32°.
a) Determine the distance, PX, across the pond, b) Determine the distance between point X and
rounded to the nearest meter.
the person to the nearest meter.
We can use the Pythagorean Theorem:
1092  1752  y 2
or
y  206
109
175
, which will result
sin 32 
or cos32 
y
y
in y = 206 meters.
xp
175
xp  175* tan 32  109 meters
tan 32 
HW#7: P308 – 309: 1, 3, 5, 25, 26
P314 – 315: 1, 3, 5, 14, 15, 16.
Solutions
P308-309:
1) x = 13.7
3) x = 48.3
5) x = 55.4
25) a) 0.7002; 0.4663; 1.1665
b) 60; 1.7321
c) no
d) no; tan35 – tan 25 = 0.2339 while tan (35-25) = 0.1763
p
q
5
8
26) a) tan P  and tan Q  so tan P  tan Q  1
b) 58  90  32 ,  tan 58  1 so tan 32 
q
p
8
5
P314-315:
1) x = 21; y = 28
3) x = 89; y = 64
15) AB = 149 m
16) x = 350 m
5) x =28; y= 10
14) x = 83