Download Unit 7b

Right Triangles and
Trigonometry
By Mathew Chacko and Jessica Wu
illuminati
Overview of Concepts
● Sine, Cosine, and Tangent help provide ratios that
find the interior angles and side lengths of triangles.
● SOHCAHTOA use when right triangle
● first, you must use an angle that is not right (90)
● determine whether to use sin, cos, or tan based on
the side lengths you know (opp and hyp=sin, adj and
hyp=cos, adj and opp=tan)
● put sin,cos, or tan in front of said angle measure and
set equal to trig ratio
Inverse SOHCAHTOA
● find a proportion using unknown angle and trig ratios (sin, cos, tan)
● tan(A)=opp/adj, sin(A)=opp/hyp, cos(A)=adj/hyp
● if using sine, find the opp and hyp of said angle, same for cosine and
tangent
● after finding a fraction using given numbers, apply inverse to the
sign
● example below (easy)
find measure of <A
use tan b/c you know opp
and adj side compared to
<A
tan(A)=16/20=4/5
apply inverse
tan-1(4/5) is about 38.7
<A≈38.7 degrees
16
A
20
Law of Sine
A,B,C are angle measurements
a,b,c are side lengths
law of sine is sinA/a= sinB/b= sinC/c
first proportion you use must have given a side and angle
proportion you are solving for must have only one known side or
angle
● angle measure over side set equal to angle/side (unknown)
● then cross multiply to isolate unknown variable then solve
● remember it only finds acute angles
●
●
●
●
●
Law of Cosine
● law of cosine is a²=b²+c²-2bc cos(A), b²=a²+c²-2ac cos (B), c²=a²+b²2ab cos (C)
● one side squared set equal to other two sides squared minus two times
those sides multiplied together than cosine of angle corresponding with
original side
● only need to use if only know all the sides or SAS
● example below (medium)
FIND THE MISSING SIDE
a²=64+100-2(80)cos(60)
164-80=84
a²=2√21 ≈9.2
a=9.2
now that you know all the sides and an angle
you can solve the the rest of the angles using
law of sines
B
a
8
A 60°
C
10
Two Triangles, One, or None
● must test if a SSA (butt) triangle
● example, if you use sine and find an angle to be 30 degrees, you must also
test the theory using 180 minus 30 which equals 150 b/c sine only finds
acute angles
● by using side lengths, determine if one neither or both are plausible
● if a>b then A>B, if both answers work, then two triangles, if only one than
one triangle
● no solution if sine ends up greater than one
● example below (difficult)
B
a=10
c
A=24°
b=12
C
FIND THE OTHER TWO ANGLES
SSA triangle so solve for more than one possibility
sin24/10=sinB/12=sinB= sin24/10
<B≈29 degrees or 151 degrees
based on triangle, 12>10 so it could be either so two triangles
180-24-29=127, 180-151-24=6
final angle= 6 or 127 degrees
A=24 degrees B=29 or 151 degrees C=6 or 127 degrees
Connections to Other Units
● trigonometry relates to proportions b/c students make
proportions with sine, cosine and tangent
● example would be sin 62/32=sin76/b sin62(b)=sin76(32) b=33.5
● also people use SOHCAHTOA when you don’t know two sides of a
right triangle
● if two sides are given, then just apply pythagorean theorem
● which connects to unit 7a
● you may sometimes
need to use trigonometry
to find the area/volume of a triangle
in a 3-D or 2-D shape
Common Mistakes
● many students forget the ratios
for sine, cosine, and tangent a
simple and easy trick to help
memorize these confusing ratios
would be the word
“SOHCAHTOA” pic on the right
● students also forget to solve
for more than one solution when
you have a “butt” triangle or
side side angle triangle (SSA)
● remember in previous units when
we prove triangles congruent we
never use SSA so you must solve
for all possible solutions if SSA
Real Life Scenario
You are building a ramp for a moving company that slants upward at a
certain angle degree. You know that the side on the ground is 10m and the
slant upwards is 12m. You also know that the angle on the ground is 24°.
Find the rest of the dimensions.
B
Work:
1. sin24/12= sinB/10
(multiply the proportionality)
2.sin24x10= sinBx12
(use calculator to reduce equation)
a=12m
c
(use law of sine)
24°
A
C
b=10m
3.angle B is about 19.8 degreees
(not 180-19.8 b/c a>b)
4.180 - two <’s= 136.2 degrees
5.sin 24/12=sin 136.2/c
(cross multiply)
6.c is about 20.4
a=12 b=10 c=20.4
<A=24 <B=19.8 <C=
136.2