elec{cjs - wikispaces · solving for missing angles using soh cah toa . warm up ... angles of...

413

r eleccjSMrs Bellavance Name ------ _ Algebra 2Trig 1 ~ Date V rv 2 o I------=Jr--=-=-c-----shy

Right Triangle Trigonometry Practice Remember - Soh Cah Toa

Looking Back at the 30deg 60deg 90deg

sm S c)

t

cos

--

XJ3 h-- j 2x

~~

tan - _~

shy shy

lc~XJ3 330deg

I

~3Z x f 2shy

60deg J3- shy

J L 53

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

45 ~ - shy2shy

sshy- shy

L I

r ----shy

E Jl

Practice - Write the 3 trigonoCfjc 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 5sinB = I~reg

(cosB= 10

813ltD tanB= ~ oGJ

b2+- C-=Io -a

_ Vi F B

L-

1 7 Ji

2M cosB

6 B 1 3 sinB sm =~ shy

0J1- shy

6 i cos B = i- shy-a 5~

tanBtanO~ f

6 5Sino= (jJsmB= shy3

8-16 ~COSO~reg

COSB=iJi) ~ 12

tano=(]j 13

tanO~ cJj)(2

I~ c 1= S- 1- 17 13 1_ F--- 62 - OYCo 8J1 Vi Vi 2- Cgt I t1

b co ~ ~h c ~715 ~~ I

-z

--

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

22shy

lj ------~

~ ~

x ~

r)(~

-

~=Lftgtmiddot~~l

22 I-It

Y N-

x X 2705 1 (n)

rX ~ 17S7l

X ( 0)- -c gt1 s~ 27

--

Opp

Y

I~ f p

X

j - 12 t clt r (2 Z )

r1~ Lf~ gS- ] -

_1_2_ = X Cogt(n)

rx = 12~

9

1 s I ( 2 s)

380 I J -== I Co S ( 25)

r l~-= RmiddotIG -------shy

10

x

~ ~ ~ t-c ( ~ )

r~ ~ 703

X -=shy 11lL 11

47 y rAi

12

rX II omiddot Lf 10 (

o X

17-0

~ -shy 10 C2JSc b 7 )

~~-

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

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

Mrs BeIJavance Name Algebra 2Trig 1 Block Date

131 - Right Triangle Trigonometry

Trigonometry is a Greek term meaning t-r- Cilt J I~ tv e ~s ~~r ( -L-+

Triangles consist of two components S 0 U

RatiosRelations in triangles dependent on the r e+e( f ( L- aIIj I L

()

e

The ratios formed by the sides have nameslabels

sin B=

cos B =

tan B=

_0 ffv~Ik

1- Yfgt ~~ IA stshy

A)) I-CtA fshy

HypD k01 ltshy

degUJ5 k-Adjvl~-t-

- middot middot Remembermiddot middot middot middot middot middot c~middot middot Hmiddot middot middot middot middot middot middotbullbullbullbullbullbullbullbullbullbullbullbullbullbullbull~

Example Write 3 basic trigonometric ratios for the following triangles

s sin B= -Y

5 -I cos B= 3 S-

sin B= gshy17

8

cos B= (~

7

tan B= 3-1

L z 81 + IS ltshy

tan B= 8 ~

o -I Is-

C - ~z tshy (S

4 L 7

-------

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 ~yOdeg 1deg y f

oJj _

X ope

X -= lOs (70 0

)

IX z 9 LI rJ - ( 0 S (70 )

0

CJ = 0 (gt~ hlt)

J -= ~ L)

3

9 x

y

~ - s (25) c

0- 151-(25)

2 r x~90~It off

13 q)

)(

3

X - 3 ttlt1 (19)

rx - Lfmiddot ~I-

4 -4 ~ -t c ltQ ~ ijl1 ~ 1 pound =-- 0 A

~ - f[I (Lfb )

0-= 8ttlt(CIb)

10--0 6 3(

COS(-(b)

-= c

8

x

Mrs Bellavance Name Algebra 2Trig 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)-=- LOS67 ) = 32shy 7X -- =- ~ 5-(b)

x = 31 S~ (b 2-) j = 31 coS (6 z) 7- J( fa-(braquo

j~ 772shyIX =-7--g--~~l r~ =- Is~~~l

-=-

T ---~ 32G5= (

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 smiddotd~ (f-c-jf- L

=

sin cos tan are used when solving for the 0 j eshy

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 sCos e = 0s ( B i

gshy 6 =- COS -I (~) o e 0 h~~ -I ( ~)

- bO- 3S-5-( J Ie0

o bull( G- = 30 0J 0 0- I

13 rrJj

Iu e _9_ 3

-1 I)X 2 z 2f( +- btr e ~ tv (3-)Is

x1 ZS-V

lt ~ J 2 so ~ ~ JiO raquo = S~

3

OP( 6

lXL - b -r 71 -tc e = ---L f

36 e t4-1 ( )A 1 = -+- l( XL 8Sshy lfo-~o 7

[Ilt = J ss-

2 1 1 shy

X =8 +-12

)(1 - 6 L( i I-IL

)( - J208- JihJi3

r(~ LfJT31

e = eu- -I ( lsect )

r8 ~ s c 31 ~ l 4

~2oPP 9 ~LJ

x

2X- - 2 2 ~ S ~ $ e- shy s )(l + - 7-5

-= Sfgt -I ( )e y z 2 I

-re - zsss 0 JXgt-~

5 6

Opp 9

-fi

LCos G- 12 (AcJJ

e- t) XL-= 9+-(21C-05 -I (

Xl ~g ~~ 70_ n O

x-=~

X JY J2

~ii--l

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

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

LR~~ 2YAJ3 7

10

Use your knowledge of306090 and 454590 special right triangles to find the missing sides -XJ3

3 0 2J3 4 M

30deg 312 )

2 2 -- -X2 (2) --=t-shy

45degY 5

5 E 6 T

2 3 szshy 2h 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 () ~ SLII)X S

- ~ sshy-- j( -=c 5 (ell) cgt S (ell) -shy

L__ [lj ~

- Y SJ-]I~-G G~ I Answer

G x = 728

y = 529

E 23

2312

0

213

4

rv

3

Eshyx = 663

y = 435

-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-------------shy

I O N

x -- 53 t-cA ( ~ltt) IX-=- 277 I ____---shy---- - J3 - ~X-=Z71 (__ L1 -

CgtS (sci) ~ shy y c0s(I)

--_~

Sreg~Bse1our knowledge of trigonometry to find the missing angle

Ij - S CoS (11)

12 N ( -j = ~ -I c-r

Cos 8 ~---E bh

ft = colt-I ( 6~)5 N l

l3T 14 S 3 orr

s1 e- -= 2 (Adj

4-fi e ~ amp -I (I)7 e~ e ~ SI-() e LiSo

e -=- 2Smiddots~

Use trigonometry to answer the following real life situations T 15 A ski slope at a mountain has an angle of elevation of 52deg The length of the slope is 1600f1

How high is the mountaincJ T (

1(00~o~ Ioltf ~ I -y

52deg rl

GOO 16 Determine the angle of elevation of a plane that is traveling at an altitude of 4000 ft and is 2500 ft away (horizontal distance)

11 E

( -_-----------shy

Answer

8 ~ --M - (-~~~ ~ e = 57 or 7 shy 5 ~ cgt

-r A N G E I N T s II

X=277 I X= 227

B = 2538deg Y=327 B= 539deg 58deg B= 6465deg1 Y=446 12608 ft B=45deg I

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 aeroplanes altitude is 2OO m how far away is it bull

2500 m

Let the distance be x Then

2500 Sill 23deg = --shy

x

2500 X = = 6400rI1

sin 23deg

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 bull

highest sail ofthe Opera House

This is a simple tan ratio problem ( -tan 80 = h500

So

h = 500 tan 80 = 7027 m

So the height of the tallest point is around 70 m

[The actual height is 674 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 Lc )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 --- 51GL(-

~ AR

-middotX 0-IT

3~middotGlaquo( 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

2LI yfAlrS

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 48deg How tall is the flagpole

~ i (48)- SO

~

ex SO middoth (lf~)

148- 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 31deg 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 215 meters due west of point A and

--xis able to measure that from point C point B makes an angle of 3640 north of the line from ~ shy

~3640 nverC to A Determine the distance AB C --~------------~~ 2 rr- A

( ~ bv- (sC t ) ZIS-

Example 2 A vertical television transmission tower is 112 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 75deg with the ground Assuming the ground is horizontal in the vicinity of the base of the tower and that the support cables are stretched tight enough so that they essentially form a straight line getermine how far the cable anchors will be JillDJJhe base of the tmver and how long the cables must bR ~ijl~m~mlm~m~~llIIll==~IlI1l=1I

r tCA ( -$

111- X

h

I 3 0 0 I ~ - 1159lt)

Two boats leave the same port One goes 10 miles due west and drops anchor The other leaves the port 20 degrees

north of west How far must it go in a straight line to get as far west as the first boat

-10 - (vgt (2ltJ)

X

0)( - shyC~~(NI

X~ 100-(

A man flies a kite with a 100 foot string The angle of elevation of the string is 52 0 How high off the ground is the kite

(0 0

An airplane takes off 200 yards in front of a 60 foot building At what angle of elevation must the plane take off in order

to avoid crashing into the building Assume that the airplane flies in a straight line and the angle of elevation remains

constant until the airplane flies over the building

200 yds

GOJ p+-v

bOO ~v (~J)

X - 57 0

A 14 foot ladder is used to scale a 13 foot wall At what angle of elevation must the ladder be situated in order to reach the top of the wall

A ramp is needed to allow vehicles to climb a 2 foot wall The angle of elevation in order for the vehicles to safely go up

must be 30 0 or less and the longest ramp available is 5 feet long Can this ramp be used safely ------~-

--

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

22shy

lj ------~

~ ~

x ~

r)(~

-

~=Lftgtmiddot~~l

22 I-It

Y N-

x X 2705 1 (n)

rX ~ 17S7l

X ( 0)- -c gt1 s~ 27

--

Opp

Y

I~ f p

X

j - 12 t clt r (2 Z )

r1~ Lf~ gS- ] -

_1_2_ = X Cogt(n)

rx = 12~

9

1 s I ( 2 s)

380 I J -== I Co S ( 25)

r l~-= RmiddotIG -------shy

10

x

~ ~ ~ t-c ( ~ )

r~ ~ 703

X -=shy 11lL 11

47 y rAi

12

rX II omiddot Lf 10 (

o X

17-0

~ -shy 10 C2JSc b 7 )

~~-

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

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






()

e


sin B=

cos B =

tan B=

_0 ffv~Ik

1- Yfgt ~~ IA stshy

A)) I-CtA fshy

HypD k01 ltshy

degUJ5 k-Adjvl~-t-



s sin B= -Y

5 -I cos B= 3 S-

sin B= gshy17

8

cos B= (~

7

tan B= 3-1

L z 81 + IS ltshy

tan B= 8 ~

o -I Is-

C - ~z tshy (S

4 L 7

-------





1 ~yOdeg 1deg y f

oJj _

X ope

X -= lOs (70 0

)

IX z 9 LI rJ - ( 0 S (70 )

0

CJ = 0 (gt~ hlt)

J -= ~ L)

3

9 x

y

~ - s (25) c

0- 151-(25)

2 r x~90~It off

13 q)

)(

3

X - 3 ttlt1 (19)

rx - Lfmiddot ~I-

4 -4 ~ -t c ltQ ~ ijl1 ~ 1 pound =-- 0 A

~ - f[I (Lfb )

0-= 8ttlt(CIb)

10--0 6 3(

COS(-(b)

-= c

8

x


-~--------------

Date_-----=+---=--=-~__



y 7 teA) (6)-=- LOS67 ) = 32shy 7X -- =- ~ 5-(b)


j~ 772shyIX =-7--g--~~l r~ =- Is~~~l

-=-

T ---~ 32G5= (



=





8 8 8 10

5 5

8

4 sCos e = 0s ( B i

gshy 6 =- COS -I (~) o e 0 h~~ -I ( ~)

- bO- 3S-5-( J Ie0

o bull( G- = 30 0J 0 0- I

13 rrJj

Iu e _9_ 3

-1 I)X 2 z 2f( +- btr e ~ tv (3-)Is

x1 ZS-V


3

OP( 6

lXL - b -r 71 -tc e = ---L f

36 e t4-1 ( )A 1 = -+- l( XL 8Sshy lfo-~o 7

[Ilt = J ss-

2 1 1 shy

X =8 +-12

)(1 - 6 L( i I-IL

)( - J208- JihJi3

r(~ LfJT31


r8 ~ s c 31 ~ l 4

~2oPP 9 ~LJ

x

2X- - 2 2 ~ S ~ $ e- shy s )(l + - 7-5

-= Sfgt -I ( )e y z 2 I

-re - zsss 0 JXgt-~

5 6

Opp 9

-fi

LCos G- 12 (AcJJ

e- t) XL-= 9+-(21C-05 -I (

Xl ~g ~~ 70_ n O

x-=~

X JY J2

~ii--l


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


LR~~ 2YAJ3 7

10


3 0 2J3 4 M

30deg 312 )

2 2 -- -X2 (2) --=t-shy

45degY 5

5 E 6 T


2 S


7 G 8 E

x

s - to 5 (-I () ~ SLII)X S


L__ [lj ~


G x = 728

y = 529

E 23

2312

0

213

4

rv

3

Eshyx = 663

y = 435

-r 16 212

~

165

y 513

----

Riddle - What does trigonometry have in common with a beach


1-------------shy

I O N



--_~


Ij - S CoS (11)

12 N ( -j = ~ -I c-r

Cos 8 ~---E bh


l3T 14 S 3 orr

s1 e- -= 2 (Adj


e -=- 2Smiddots~




52deg rl


11 E

( -_-----------shy

Answer

8 ~ --M - (-~~~ ~ e = 57 or 7 shy 5 ~ cgt


X=277 I X= 227




horizontal


2500 m



x

2500 X = = 6400rI1

sin 23deg


horizontal





So

h = 500 tan 80 = 7027 m



4m

Applications


X Lc )2 I - I L

i( -- J (o-S-

- z [c 25 1


-~

tree ---~--~--~-~Y-

1 a -----

60 feet(JJ7


~ AR

-middotX 0-IT




- ssrI -- sO l2- G- 90 45D

-S - 30 z Z 1 7

~

S 30 yards

Z ( 1 IsJ1- z

45

2LI yfAlrS



~ i (48)- SO

~

ex SO middoth (lf~)

148- d

50

S-S-SJD-- C+


IQ()

b

6 100






r tCA ( -$

111- X

h

I 3 0 0 I ~ - 1159lt)



-10 - (vgt (2ltJ)

X

0)( - shyC~~(NI

X~ 100-(


(0 0




200 yds

GOJ p+-v

bOO ~v (~J)

X - 57 0




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

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






()

e


sin B=

cos B =

tan B=

_0 ffv~Ik

1- Yfgt ~~ IA stshy

A)) I-CtA fshy

HypD k01 ltshy

degUJ5 k-Adjvl~-t-



s sin B= -Y

5 -I cos B= 3 S-

sin B= gshy17

8

cos B= (~

7

tan B= 3-1

L z 81 + IS ltshy

tan B= 8 ~

o -I Is-

C - ~z tshy (S

4 L 7

-------





1 ~yOdeg 1deg y f

oJj _

X ope

X -= lOs (70 0

)

IX z 9 LI rJ - ( 0 S (70 )

0

CJ = 0 (gt~ hlt)

J -= ~ L)

3

9 x

y

~ - s (25) c

0- 151-(25)

2 r x~90~It off

13 q)

)(

3

X - 3 ttlt1 (19)

rx - Lfmiddot ~I-

4 -4 ~ -t c ltQ ~ ijl1 ~ 1 pound =-- 0 A

~ - f[I (Lfb )

0-= 8ttlt(CIb)

10--0 6 3(

COS(-(b)

-= c

8

x


-~--------------

Date_-----=+---=--=-~__



y 7 teA) (6)-=- LOS67 ) = 32shy 7X -- =- ~ 5-(b)


j~ 772shyIX =-7--g--~~l r~ =- Is~~~l

-=-

T ---~ 32G5= (



=





8 8 8 10

5 5

8

4 sCos e = 0s ( B i

gshy 6 =- COS -I (~) o e 0 h~~ -I ( ~)

- bO- 3S-5-( J Ie0

o bull( G- = 30 0J 0 0- I

13 rrJj

Iu e _9_ 3

-1 I)X 2 z 2f( +- btr e ~ tv (3-)Is

x1 ZS-V


3

OP( 6

lXL - b -r 71 -tc e = ---L f

36 e t4-1 ( )A 1 = -+- l( XL 8Sshy lfo-~o 7

[Ilt = J ss-

2 1 1 shy

X =8 +-12

)(1 - 6 L( i I-IL

)( - J208- JihJi3

r(~ LfJT31


r8 ~ s c 31 ~ l 4

~2oPP 9 ~LJ

x

2X- - 2 2 ~ S ~ $ e- shy s )(l + - 7-5

-= Sfgt -I ( )e y z 2 I

-re - zsss 0 JXgt-~

5 6

Opp 9

-fi

LCos G- 12 (AcJJ

e- t) XL-= 9+-(21C-05 -I (

Xl ~g ~~ 70_ n O

x-=~

X JY J2

~ii--l




LR~~ 2YAJ3 7

10


3 0 2J3 4 M

30deg 312 )

2 2 -- -X2 (2) --=t-shy

45degY 5

5 E 6 T


2 S


7 G 8 E

x

s - to 5 (-I () ~ SLII)X S


L__ [lj ~


G x = 728

y = 529

E 23

2312

0

213

4

rv

3

Eshyx = 663

y = 435

-r 16 212

~

165

y 513

----



1-------------shy

I O N



--_~


Ij - S CoS (11)

12 N ( -j = ~ -I c-r

Cos 8 ~---E bh


l3T 14 S 3 orr

s1 e- -= 2 (Adj


e -=- 2Smiddots~




52deg rl


11 E

( -_-----------shy

Answer

8 ~ --M - (-~~~ ~ e = 57 or 7 shy 5 ~ cgt


X=277 I X= 227




horizontal


2500 m



x

2500 X = = 6400rI1

sin 23deg


horizontal





So

h = 500 tan 80 = 7027 m



4m

Applications


X Lc )2 I - I L

i( -- J (o-S-

- z [c 25 1


-~

tree ---~--~--~-~Y-

1 a -----

60 feet(JJ7


~ AR

-middotX 0-IT




- ssrI -- sO l2- G- 90 45D

-S - 30 z Z 1 7

~

S 30 yards

Z ( 1 IsJ1- z

45

2LI yfAlrS



~ i (48)- SO

~

ex SO middoth (lf~)

148- d

50

S-S-SJD-- C+


IQ()

b

6 100






r tCA ( -$

111- X

h

I 3 0 0 I ~ - 1159lt)



-10 - (vgt (2ltJ)

X

0)( - shyC~~(NI

X~ 100-(


(0 0




200 yds

GOJ p+-v

bOO ~v (~J)

X - 57 0




-------





1 ~yOdeg 1deg y f

oJj _

X ope

X -= lOs (70 0

)

IX z 9 LI rJ - ( 0 S (70 )

0

CJ = 0 (gt~ hlt)

J -= ~ L)

3

9 x

y

~ - s (25) c

0- 151-(25)

2 r x~90~It off

13 q)

)(

3

X - 3 ttlt1 (19)

rx - Lfmiddot ~I-

4 -4 ~ -t c ltQ ~ ijl1 ~ 1 pound =-- 0 A

~ - f[I (Lfb )

0-= 8ttlt(CIb)

10--0 6 3(

COS(-(b)

-= c

8

x


-~--------------

Date_-----=+---=--=-~__



y 7 teA) (6)-=- LOS67 ) = 32shy 7X -- =- ~ 5-(b)


j~ 772shyIX =-7--g--~~l r~ =- Is~~~l

-=-

T ---~ 32G5= (



=





8 8 8 10

5 5

8

4 sCos e = 0s ( B i

gshy 6 =- COS -I (~) o e 0 h~~ -I ( ~)

- bO- 3S-5-( J Ie0

o bull( G- = 30 0J 0 0- I

13 rrJj

Iu e _9_ 3

-1 I)X 2 z 2f( +- btr e ~ tv (3-)Is

x1 ZS-V


3

OP( 6

lXL - b -r 71 -tc e = ---L f

36 e t4-1 ( )A 1 = -+- l( XL 8Sshy lfo-~o 7

[Ilt = J ss-

2 1 1 shy

X =8 +-12

)(1 - 6 L( i I-IL

)( - J208- JihJi3

r(~ LfJT31


r8 ~ s c 31 ~ l 4

~2oPP 9 ~LJ

x

2X- - 2 2 ~ S ~ $ e- shy s )(l + - 7-5

-= Sfgt -I ( )e y z 2 I

-re - zsss 0 JXgt-~

5 6

Opp 9

-fi

LCos G- 12 (AcJJ

e- t) XL-= 9+-(21C-05 -I (

Xl ~g ~~ 70_ n O

x-=~

X JY J2

~ii--l




LR~~ 2YAJ3 7

10


3 0 2J3 4 M

30deg 312 )

2 2 -- -X2 (2) --=t-shy

45degY 5

5 E 6 T


2 S


7 G 8 E

x

s - to 5 (-I () ~ SLII)X S


L__ [lj ~


G x = 728

y = 529

E 23

2312

0

213

4

rv

3

Eshyx = 663

y = 435

-r 16 212

~

165

y 513

----



1-------------shy

I O N



--_~


Ij - S CoS (11)

12 N ( -j = ~ -I c-r

Cos 8 ~---E bh


l3T 14 S 3 orr

s1 e- -= 2 (Adj


e -=- 2Smiddots~




52deg rl


11 E

( -_-----------shy

Answer

8 ~ --M - (-~~~ ~ e = 57 or 7 shy 5 ~ cgt


X=277 I X= 227




horizontal


2500 m



x

2500 X = = 6400rI1

sin 23deg


horizontal





So

h = 500 tan 80 = 7027 m



4m

Applications


X Lc )2 I - I L

i( -- J (o-S-

- z [c 25 1


-~

tree ---~--~--~-~Y-

1 a -----

60 feet(JJ7


~ AR

-middotX 0-IT




- ssrI -- sO l2- G- 90 45D

-S - 30 z Z 1 7

~

S 30 yards

Z ( 1 IsJ1- z

45

2LI yfAlrS



~ i (48)- SO

~

ex SO middoth (lf~)

148- d

50

S-S-SJD-- C+


IQ()

b

6 100






r tCA ( -$

111- X

h

I 3 0 0 I ~ - 1159lt)



-10 - (vgt (2ltJ)

X

0)( - shyC~~(NI

X~ 100-(


(0 0




200 yds

GOJ p+-v

bOO ~v (~J)

X - 57 0





-~--------------

Date_-----=+---=--=-~__



y 7 teA) (6)-=- LOS67 ) = 32shy 7X -- =- ~ 5-(b)


j~ 772shyIX =-7--g--~~l r~ =- Is~~~l

-=-

T ---~ 32G5= (



=





8 8 8 10

5 5

8

4 sCos e = 0s ( B i

gshy 6 =- COS -I (~) o e 0 h~~ -I ( ~)

- bO- 3S-5-( J Ie0

o bull( G- = 30 0J 0 0- I

13 rrJj

Iu e _9_ 3

-1 I)X 2 z 2f( +- btr e ~ tv (3-)Is

x1 ZS-V


3

OP( 6

lXL - b -r 71 -tc e = ---L f

36 e t4-1 ( )A 1 = -+- l( XL 8Sshy lfo-~o 7

[Ilt = J ss-

2 1 1 shy

X =8 +-12

)(1 - 6 L( i I-IL

)( - J208- JihJi3

r(~ LfJT31


r8 ~ s c 31 ~ l 4

~2oPP 9 ~LJ

x

2X- - 2 2 ~ S ~ $ e- shy s )(l + - 7-5

-= Sfgt -I ( )e y z 2 I

-re - zsss 0 JXgt-~

5 6

Opp 9

-fi

LCos G- 12 (AcJJ

e- t) XL-= 9+-(21C-05 -I (

Xl ~g ~~ 70_ n O

x-=~

X JY J2

~ii--l




LR~~ 2YAJ3 7

10


3 0 2J3 4 M

30deg 312 )

2 2 -- -X2 (2) --=t-shy

45degY 5

5 E 6 T


2 S


7 G 8 E

x

s - to 5 (-I () ~ SLII)X S


L__ [lj ~


G x = 728

y = 529

E 23

2312

0

213

4

rv

3

Eshyx = 663

y = 435

-r 16 212

~

165

y 513

----



1-------------shy

I O N



--_~


Ij - S CoS (11)

12 N ( -j = ~ -I c-r

Cos 8 ~---E bh


l3T 14 S 3 orr

s1 e- -= 2 (Adj


e -=- 2Smiddots~




52deg rl


11 E

( -_-----------shy

Answer

8 ~ --M - (-~~~ ~ e = 57 or 7 shy 5 ~ cgt


X=277 I X= 227




horizontal


2500 m



x

2500 X = = 6400rI1

sin 23deg


horizontal





So

h = 500 tan 80 = 7027 m



4m

Applications


X Lc )2 I - I L

i( -- J (o-S-

- z [c 25 1


-~

tree ---~--~--~-~Y-

1 a -----

60 feet(JJ7


~ AR

-middotX 0-IT




- ssrI -- sO l2- G- 90 45D

-S - 30 z Z 1 7

~

S 30 yards

Z ( 1 IsJ1- z

45

2LI yfAlrS



~ i (48)- SO

~

ex SO middoth (lf~)

148- d

50

S-S-SJD-- C+


IQ()

b

6 100






r tCA ( -$

111- X

h

I 3 0 0 I ~ - 1159lt)



-10 - (vgt (2ltJ)

X

0)( - shyC~~(NI

X~ 100-(


(0 0




200 yds

GOJ p+-v

bOO ~v (~J)

X - 57 0




13 rrJj

Iu e _9_ 3

-1 I)X 2 z 2f( +- btr e ~ tv (3-)Is

x1 ZS-V


3

OP( 6

lXL - b -r 71 -tc e = ---L f

36 e t4-1 ( )A 1 = -+- l( XL 8Sshy lfo-~o 7

[Ilt = J ss-

2 1 1 shy

X =8 +-12

)(1 - 6 L( i I-IL

)( - J208- JihJi3

r(~ LfJT31


r8 ~ s c 31 ~ l 4

~2oPP 9 ~LJ

x

2X- - 2 2 ~ S ~ $ e- shy s )(l + - 7-5

-= Sfgt -I ( )e y z 2 I

-re - zsss 0 JXgt-~

5 6

Opp 9

-fi

LCos G- 12 (AcJJ

e- t) XL-= 9+-(21C-05 -I (

Xl ~g ~~ 70_ n O

x-=~

X JY J2

~ii--l




LR~~ 2YAJ3 7

10


3 0 2J3 4 M

30deg 312 )

2 2 -- -X2 (2) --=t-shy

45degY 5

5 E 6 T


2 S


7 G 8 E

x

s - to 5 (-I () ~ SLII)X S


L__ [lj ~


G x = 728

y = 529

E 23

2312

0

213

4

rv

3

Eshyx = 663

y = 435

-r 16 212

~

165

y 513

----



1-------------shy

I O N



--_~


Ij - S CoS (11)

12 N ( -j = ~ -I c-r

Cos 8 ~---E bh


l3T 14 S 3 orr

s1 e- -= 2 (Adj


e -=- 2Smiddots~




52deg rl


11 E

( -_-----------shy

Answer

8 ~ --M - (-~~~ ~ e = 57 or 7 shy 5 ~ cgt


X=277 I X= 227




horizontal


2500 m



x

2500 X = = 6400rI1

sin 23deg


horizontal





So

h = 500 tan 80 = 7027 m



4m

Applications


X Lc )2 I - I L

i( -- J (o-S-

- z [c 25 1


-~

tree ---~--~--~-~Y-

1 a -----

60 feet(JJ7


~ AR

-middotX 0-IT




- ssrI -- sO l2- G- 90 45D

-S - 30 z Z 1 7

~

S 30 yards

Z ( 1 IsJ1- z

45

2LI yfAlrS



~ i (48)- SO

~

ex SO middoth (lf~)

148- d

50

S-S-SJD-- C+


IQ()

b

6 100






r tCA ( -$

111- X

h

I 3 0 0 I ~ - 1159lt)



-10 - (vgt (2ltJ)

X

0)( - shyC~~(NI

X~ 100-(


(0 0




200 yds

GOJ p+-v

bOO ~v (~J)

X - 57 0







LR~~ 2YAJ3 7

10


3 0 2J3 4 M

30deg 312 )

2 2 -- -X2 (2) --=t-shy

45degY 5

5 E 6 T


2 S


7 G 8 E

x

s - to 5 (-I () ~ SLII)X S


L__ [lj ~


G x = 728

y = 529

E 23

2312

0

213

4

rv

3

Eshyx = 663

y = 435

-r 16 212

~

165

y 513

----



1-------------shy

I O N



--_~


Ij - S CoS (11)

12 N ( -j = ~ -I c-r

Cos 8 ~---E bh


l3T 14 S 3 orr

s1 e- -= 2 (Adj


e -=- 2Smiddots~




52deg rl


11 E

( -_-----------shy

Answer

8 ~ --M - (-~~~ ~ e = 57 or 7 shy 5 ~ cgt


X=277 I X= 227




horizontal


2500 m



x

2500 X = = 6400rI1

sin 23deg


horizontal





So

h = 500 tan 80 = 7027 m



4m

Applications


X Lc )2 I - I L

i( -- J (o-S-

- z [c 25 1


-~

tree ---~--~--~-~Y-

1 a -----

60 feet(JJ7


~ AR

-middotX 0-IT




- ssrI -- sO l2- G- 90 45D

-S - 30 z Z 1 7

~

S 30 yards

Z ( 1 IsJ1- z

45

2LI yfAlrS



~ i (48)- SO

~

ex SO middoth (lf~)

148- d

50

S-S-SJD-- C+


IQ()

b

6 100






r tCA ( -$

111- X

h

I 3 0 0 I ~ - 1159lt)



-10 - (vgt (2ltJ)

X

0)( - shyC~~(NI

X~ 100-(


(0 0




200 yds

GOJ p+-v

bOO ~v (~J)

X - 57 0




----



1-------------shy

I O N



--_~


Ij - S CoS (11)

12 N ( -j = ~ -I c-r

Cos 8 ~---E bh


l3T 14 S 3 orr

s1 e- -= 2 (Adj


e -=- 2Smiddots~




52deg rl


11 E

( -_-----------shy

Answer

8 ~ --M - (-~~~ ~ e = 57 or 7 shy 5 ~ cgt


X=277 I X= 227




horizontal


2500 m



x

2500 X = = 6400rI1

sin 23deg


horizontal





So

h = 500 tan 80 = 7027 m



4m

Applications


X Lc )2 I - I L

i( -- J (o-S-

- z [c 25 1


-~

tree ---~--~--~-~Y-

1 a -----

60 feet(JJ7


~ AR

-middotX 0-IT




- ssrI -- sO l2- G- 90 45D

-S - 30 z Z 1 7

~

S 30 yards

Z ( 1 IsJ1- z

45

2LI yfAlrS



~ i (48)- SO

~

ex SO middoth (lf~)

148- d

50

S-S-SJD-- C+


IQ()

b

6 100






r tCA ( -$

111- X

h

I 3 0 0 I ~ - 1159lt)



-10 - (vgt (2ltJ)

X

0)( - shyC~~(NI

X~ 100-(


(0 0




200 yds

GOJ p+-v

bOO ~v (~J)

X - 57 0






horizontal


2500 m



x

2500 X = = 6400rI1

sin 23deg


horizontal





So

h = 500 tan 80 = 7027 m



4m

Applications


X Lc )2 I - I L

i( -- J (o-S-

- z [c 25 1


-~

tree ---~--~--~-~Y-

1 a -----

60 feet(JJ7


~ AR

-middotX 0-IT




- ssrI -- sO l2- G- 90 45D

-S - 30 z Z 1 7

~

S 30 yards

Z ( 1 IsJ1- z

45

2LI yfAlrS



~ i (48)- SO

~

ex SO middoth (lf~)

148- d

50

S-S-SJD-- C+


IQ()

b

6 100






r tCA ( -$

111- X

h

I 3 0 0 I ~ - 1159lt)



-10 - (vgt (2ltJ)

X

0)( - shyC~~(NI

X~ 100-(


(0 0




200 yds

GOJ p+-v

bOO ~v (~J)

X - 57 0





horizontal





So

h = 500 tan 80 = 7027 m



4m

Applications


X Lc )2 I - I L

i( -- J (o-S-

- z [c 25 1


-~

tree ---~--~--~-~Y-

1 a -----

60 feet(JJ7


~ AR

-middotX 0-IT




- ssrI -- sO l2- G- 90 45D

-S - 30 z Z 1 7

~

S 30 yards

Z ( 1 IsJ1- z

45

2LI yfAlrS



~ i (48)- SO

~

ex SO middoth (lf~)

148- d

50

S-S-SJD-- C+


IQ()

b

6 100






r tCA ( -$

111- X

h

I 3 0 0 I ~ - 1159lt)



-10 - (vgt (2ltJ)

X

0)( - shyC~~(NI

X~ 100-(


(0 0




200 yds

GOJ p+-v

bOO ~v (~J)

X - 57 0




4m

Applications


X Lc )2 I - I L

i( -- J (o-S-

- z [c 25 1


-~

tree ---~--~--~-~Y-

1 a -----

60 feet(JJ7


~ AR

-middotX 0-IT




- ssrI -- sO l2- G- 90 45D

-S - 30 z Z 1 7

~

S 30 yards

Z ( 1 IsJ1- z

45

2LI yfAlrS



~ i (48)- SO

~

ex SO middoth (lf~)

148- d

50

S-S-SJD-- C+


IQ()

b

6 100






r tCA ( -$

111- X

h

I 3 0 0 I ~ - 1159lt)



-10 - (vgt (2ltJ)

X

0)( - shyC~~(NI

X~ 100-(


(0 0




200 yds

GOJ p+-v

bOO ~v (~J)

X - 57 0






- ssrI -- sO l2- G- 90 45D

-S - 30 z Z 1 7

~

S 30 yards

Z ( 1 IsJ1- z

45

2LI yfAlrS



~ i (48)- SO

~

ex SO middoth (lf~)

148- d

50

S-S-SJD-- C+


IQ()

b

6 100






r tCA ( -$

111- X

h

I 3 0 0 I ~ - 1159lt)



-10 - (vgt (2ltJ)

X

0)( - shyC~~(NI

X~ 100-(


(0 0




200 yds

GOJ p+-v

bOO ~v (~J)

X - 57 0





~ i (48)- SO

~

ex SO middoth (lf~)

148- d

50

S-S-SJD-- C+


IQ()

b

6 100






r tCA ( -$

111- X

h

I 3 0 0 I ~ - 1159lt)



-10 - (vgt (2ltJ)

X

0)( - shyC~~(NI

X~ 100-(


(0 0




200 yds

GOJ p+-v

bOO ~v (~J)

X - 57 0









r tCA ( -$

111- X

h

I 3 0 0 I ~ - 1159lt)



-10 - (vgt (2ltJ)

X

0)( - shyC~~(NI

X~ 100-(


(0 0




200 yds

GOJ p+-v

bOO ~v (~J)

X - 57 0






-10 - (vgt (2ltJ)

X

0)( - shyC~~(NI

X~ 100-(


(0 0




200 yds

GOJ p+-v

bOO ~v (~J)

X - 57 0




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