# elec{cjs - wikispaces .solving for missing angles using soh cah toa . warm up ... angles of...     Post on 26-Jun-2018

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r elec{cj"SMrs. Bellavance Name: -,----- _ Algebra 2/Trig 1 ~ Date: /V'. r;v...... '2 o : I---'---=J.::r--=-=-.c:.-----

Right Triangle Trigonometry Practice Remember - Soh Cah Toa!

Looking Back at the 30, 60, 90

sm S c)

t;

cos

/;-'-"

/ ' 'XJ3 /h-- j .' 2.x

~~

tan - ...._~

/ "

/ ,

lc~XJ3 330 I .",

~3Z x f 2.\

60 J3- ?

J. ""L 53

Looking Back at the 45, 45, 90 sm cos tan

45 ~ - 2.

s;-

L I

.r;: ; ----

\E Jl:

Practice - Write the 3 trigonoCfj...c rations for the angle B in each right triangle. 1. . B 11J3 G 2.

SIn = -'- :: :::!.2. 8 '2.

8 cos B = 1 ~ (0 6V V; {7!)

4

4. 5.sinB = I~

(,cosB= 10

813

• --

Finding Missing Sides in a Right Triangle Using Soh Cah Toa. Round to 2 decimal places. 7. 8.

2'2

lj ::: ------~ {

\ ~ ~

x ~

r)(~

-

\~=Lft>~~l

22 I-\'It

Y N-;

x X :::. 2705 1..... (n)

rX ~ 17.S7l

'X . ( 0)- -c, ::>'''"1 s~ 27

--

Opp

Y

I~ 'f p

X

!j -;. 12 t c< r, (2 Z. )

r1~ Lf~ gS- ] -

_1_2_ = 'X Co.>(n)

rx = 12~ 9.

1 s I'''' ( 2. s)

3.8"0 I :J -== I Co S ( 25)

r: l~-= RIG -------

10.

x

~ ~ ~ t-c,,", (,? ~ )

r~ ~ 7,03

'X -= 11.'lL 11.

47" y {\rAi

12.

r'X::: II o Lf 10 (

o{:' ...,, \ X

17.-0

~ -..: !10 C2JSc b 7 ) ~~-

• ----------------

--- -----------

Mrs. BeIJavance Name: Algebra 2/Trig 1 Block: Date:

13.1 - Right Triangle Trigonometry

Trigonometry is a Greek term meaning t-r-,' Ci< J I~ tv' e. ~s ~~r ( """-L-..+" Triangles consist of two components

.:. S ,'0{ U

Ratios/Relations in triangles dependent on the r e+e( f"" ( 'L- a.IIj I L

()

e

The ratios formed by the sides have names/labels:

sin B=

cos B =

tan B=

_0 ff'v~I'k

1-\ Yf.:> ~{~ IA st.

A,)...) I-\C,t.A f

HypD k"0.1

• -------

Solving for missing sides Steps: 1. Label your sides based on the angle given (opp, adj, hyp)

2. Pick one side to solve for first 3. Pick the correct trig function using SOHCAHTOA

a. Using given side and the side you are trying to find 4. Set up your equation 5. Solve (get the variable by itself)

Find the missing sides of these right triangles using trigonometry.

1. ~yO 1 "y f o.J.j _

X ope

'X -= lOs ,'", (70 0 )

IX z: 9. LI r':::J -:: ( 0 S (70 ) /0

C:J := /0 (.;;>~ h

• Mrs. Bellavance Name: Algebra 2/Trig I ~

-~----,.----------

Date:_--,---",=+---=--=-~__

Solving for Missing Angles using Soh Cah Toa

Warm Up - Find the missing sides in the right triangles. 1. 2.

y 7 teA') (6'/)-=- LOS{67. ) = 3"2 7'X -- =- ~ 5,'..-,(b:;)

x = 31. S"~ (b 2.-') !j = 31. c..oS (6 z") 7- J( fa-.(b'

j~ 7.72IX :=:-7--g--.~~l r~ =- Is~'~~l

-=-

T \ ---~ 3.'2G.5= (

Objective - To solve for the missing acute angles in a right triangle.

Solving for Missing Angles .:. sin, cos, tan are used when solving for a s./d~ (f...-,c-jf.- L,

=

.:. sin", cos', tan" are used when solving for the 0. /\,j {e...

sin (angle) = ratio sin" (ratio) = angle cos (angle) = ratio cos" (ratio) = angle tan (angle) = ratio tan -1 (ratio) = angle

Steps: 1. Label your sides bases on the angle given (opp, ad], hyp) 2. Pick the correct trig function using SOHCAHTOA 3. Set up the equation, using the inverse function. 4. Use your calculator to find the measure of the angle. 5. Round your answer to the second decimal place and use the degree symbol.

Determine the missing angles. 1. 2. 3.

8 8 8 10

5 5

8

4 s:C.os e = /0.s ,'("\ B::: 'i

g 6 =- COS -I (,~) o e ::0 h~~ -I ( ~)

- bO"- 3S-.5"'-( J Ie0 o ( G- = 30 0J 0 0- I

• 13 !'rrJj

I:u/\ e ;:::: _9_ /'3.

-1 I')X 2 z: 2f( +- /bt:r e ~ tv (3-)Is

x'1. ':. 'ZS-V

?< ~ J 2 so ~ ~ JiO ': = S~

3.

OP('\ 6

'l..'X"L -::: b -r 7'1. -tc.,.., e := ---.L f

36 e ::: t4,,-1 ( ;)A 1. = -+- l(' X'L :. 8S @: lfo-~o 7

[I< = J s?s- \

2. '1. 1. "

X =8 +-12

,)('1. -::: 6 L( i: I'-IL/

)( -;: J208'- ::: JihJi3 r?(~ LfJT31

e := eu-. -I ( l ) r8 ~ s c . 31 ~ l 4.

~2oPP : ... 9. ~LJ

x

2X"- }- 2 2 ~ S ~ \$,.", e- ::. s: ')('l. + '-/ "" 7.-5"'

-=: Sf>, -I ( ; )e y ..... z: 2. I

-re - zs.ss 0 J'X>-~

5. 6.

Opp 9

-fi

..LCo.s G-::. 12 ('A.cJJ:?

e- :::: t) XL-=: 9""+-('21.C-05 -I ( X'l.. ~g ~~ 70_ n O; x-:=:~

X ::: JY J2:

~ii--l

• ----------------Algebra 2 / Trig 1 Name Review: Right Triangles and Trigonometry Block Date _

Page 1 Riddle: What kind of tree does a math teacher climb?

Use the Pythagorean Theorem to find the missing side. Simplify answers, leave in radical form.

L"R~~ 2."Y/A\J3 7

10

Use your knowledge of30,60,90 and 45,45,90 special right triangles to find the missing sides. -XJ3

3. "0" 2J3 4. "M" 30

312 ").::2 2 -- -X2' (2)

;:-.-=t-45Y

:5

5. "E" 6. "T"

2 3 s:z: 2.h SJ ~ri ~ J(:

2 "S

Use your knowledge of trigonometry to find the missing sides. Round to 2 decimal places.

7. "G" 8. "E"

x

s: -::: to 5 (-I () ~ :: S,',,{LII)X S

- ~ s-- j( -:=c 5" ..... (ell) c..::> S (ell) ,...-.L__ [lj

~

- Y SJ-]I;;:~-G. G~ I Answer:

G x = 7.28

y = 5.29

E 23

2312

0

213

4

rv\

3

Ex = 6.63

y = 4.35

-r 16 212

~

165

y 513

• ----

Page 2 Riddle - What does trigonometry have in common with a beach?

Find the missing sides. Round to two decimal places. 9. "A" x 0

1-------------

I O. "N"

'x --.:: 53 t-c.A. ( ~

• Angles of Elevation and Depression

In surveying, the angle of elevation is the angle from the horizontal looking up to some object:

horizontal

The angle .of elevation of an aeroplane is 230 If the aeroplane's altitude is 2?OO m, how far away is it?

2500 m

Let the distance be x. Then

. 2500 Sill 23 = --

x

2500 X = = 6400rI1

sin 23

• The angle of depression is the angle from the horizontal looking down to some object:

horizontal

You can walk across the Sydney Harbour Bridge and take a photo of the Opera House from about the same height as top of the highest sail.

This photo was taken from a point about 500 m horizontally from the Opera House and we observe the waterline below the highest sail as having an angle of depression of 80 How high above sea level is the

highest sail ofthe Opera House?

This is a simple tan ratio problem. ( -,tan 80 = h/500

So

h = 500 tan 80 = 70.27 m.

So the height of the tallest point is around 70 m.

[The actual height is 67.4 m.]

• 4m

Applications

How far up a wall will an 11 m ladder reach, if the foot of the ladder must be 4m from the base of the wall?

X L.:c )'2 I - I L.

i( -- J (o-S- '

?'- z: [c: 25 "1

A tree casts a 60 foot shadow. The angle of elevation is 300. This is the angle at which you look up to the top of the tree from the ground. What is the height of the tree?

"'-~

tree --'-~--.............~--~-~Y-

'1 a -----.... 60 feet

({JJ7

t re e -== GO rv CX~ - reo ,--.- 5'1.GL(-~ \A\R

-:::..X '0-\IT 3~G( t~X--=

• A brick pathway is 30 yards long. A square courtyard is being built incorporating this pathway diagonally. How long should each side of the square be?

A diagram showing the square and the required pathway is shown below.

- ssrI -- sO \l2- '.G- 90" 45D

-S - 30 z: 'Z 1. 7.. ! ~

S 30 yards

'Z. ( '1 \ IsJ1- z:

45"

2/.'LI yfAlr/S

Looking down from the roof of a house at an angle of 230 a shiny object is seen. This 230 angle is with respect to the horizontal (see diagram below). The roof of the house is 32 feet above the ground. How far is the shiny object from the house?

• A flagpole stands in the middle of a flat, level field. Fifty feet away from its base a surveyor measures the angle to the top of the flagpole as 48. How tall is the flagpole?

~ i.: (48)-' SO ~

ex SO h.", (l.f~) 1\48- d

:::.

50

:::. S-S-.SJD-- C+.

A 100 foot wharf sits along the bank of a river. A surveyor stands directly across the river from one end of the wharf From where he stands the angle between the lines of sight to the two ends of the wharf is 31. How wide is the river?

IQ()

b

6 100

• Example 1: Hank needs to determine tile distance across a wide river in a north-south direction between points A and B, as indicated in the sketch to the right. He locates point C, which is 2'15 meters due west of point A, and

--x;is able to measure that from point C, point B makes an angle of 36.40 north of the line from ~" ./' ~36.40 nverC to A. Determine the distance AB. C ..--~------------~~.

"2 /r:r- A

/( ~ bv-/\ (sC? 't .) ZIS-

Example 2: A vertical television transmission tower is '1'12 m high. Support cables attached to its top are to be anchored in the ground at points located so that the support wire makes an angle of 75 with the ground. Assuming the ground is horizontal in the vicinity of the base of the tower and that the support ca

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