Math 10

3.3 Notes

Angles of Elevationl Depression & Solving Right Triangles

Part 1- Angles of Elevation and Depression

Angle of elevation: is the angle between the line of sight and the horizontal line(above the horizontal line).

- When a person looks at something _~"'£.i.lo'o.L.O.:-v.::..-..:::.0~__ his or herlocation.

Angle of depression: is the angle between the line of sight and the horizontal line(below the horizontal line).

- When a person looks at something __ b~\0 vJlocation.

his or her

A right triangle can be formed in these problems since there is a horizontal line, a

vertical line and a line of sight (diagonal line).

1

Things to notice:

fY\1'&S\(\j L

:::q a o - Lo~ cAi.p .

L D~ -e.\-{~Ct~ion

L -e\--e.\l D-\ \0(\ - L d~pv-t?s S I" Df!

Example: Find the angle of elevation:

A bird watcher sets up a camera on the ground and it points to the top of atree. The tree is 10m tall and the camera is set up 15 m from the base ofthe tree. What is the angle of elevation?

Of>P tree

10 m~ e ~ CGi\'Y\<'("iA .

~15m~~\j'

'~( (\c\ L e \-e. --J {,\ ~ i0(""\(e):

'- L l.\ \::).e.. \ -\y\' C-\0j \ e: .- w~ \ZV\ovJ .2ft-' 0\(\{j~,Q-cli

+-C\V'\ e-~ \0 ~ 0.\01- 2

)$ ------~9':: i-C\~ - \ (o ,lo1-) =~~ __.1-' a 1

Example: Angle of depression:

A boat is 75 m from the base of a light house. The angle of elevation fromthe boat to the top of the light house is 39.1°.

L ~5 rn ~ bOtA'\-What is the angle of depression? 3q , \ 0

What is the angle formed between the light house and the diagonalline of sight? 50, c,o q 0° _ 3<1. \ 'Q = 50 \10

How tall is the lighthouse (to the nearest metre)?lA ~-L L o~ e.\-{v t1\ -\--; D (\ Q'IS bUr

"o..(\~\0 0\- (I\-\<.("es,-\- .

- \Ct~\ t-vl'tl,(lj\(;

+0.(\ ~q. \ =- orr15

-\-0-(\ 3q·\ 1-- ,5

b /6\1.6+ i- l 5

-::: C) f P

= \ to °3,q 5\

L\ !j\-.\\lO\.AS~ \ ~ _{o_, _\_{Y\_ +C\ \ \ .

Example: Find the angle of elevation

A person stands 1.7 m above the ground and is looking at the top of atree which is 27.5 m away on level ground. The tree is 18.6 m high. Findthe angle of elevation. ( L~\'Io"~ I 1-- ,I rn

\t-

\lo\~\~)OfP \'6.b(Y\

Loo""'. ('\j +Or 8:~~ ~ "" C\\lu 2f 'P; .£.-,1j

+-Od\ e=-. 0f~d-c\j

+etv\ e ~_\b I C\ ':::' b. ~\q 5~l·~

G ~ -~'CA(\ - \ ~ o. to Pt 5 JlO 3 L ~. \

4

Example: Find the angle of elevation

Sean uses a device called a transit to measure the height of a totem pole.He positions his transit 19.0 m from the totem pole and records the angle ofelevation to the top of the totem pole to be 63°. If Sean's transit is 1.7 mhigh, how tall is the totem pole?

~\-t.t()

lOfr') fo\d

_ \ lA'oi- \ hi' anj\ (.;

\t\C~\I~ .0- ct,,) tA n. A L

\f\j lln-\- QfP

So~ CArt @

\ ,ill) 1-hrC~nSlt-

{:-----~-=>\q.D (y\ 7

l oJ,} ')

Iq.o :;.+nil ~d=.~ f y( o)f\<'O

\q,o Y' -\-·Gt()103 - DfP

\ q .0 i- v , '\ 102 ~ == V-~-'l-,-'-:'-""'----1

'To-km po\{ = 0rp "-t \: f rI\

31'~ ~ \pl ~ =-l3~~ \-\--6t\\

Last example: Calculate a distance using the angle of depression:

Natalie is rock climbing and Aaron is belaying. When Aaron pulls the rope taut tothe ground, the angle of depression is 73°. If Aaron is standing 8 ft from the wall,

what length is of rope is off the ground?

'?Jt. h'!Jp·

The angle that the rope makes with the vertical is \1 C

Find the length II' of the rope: Decide which trig ratio you will use.Le,\ \~5 \ASL \10 (AS Du.( Cl(\8 \( 0+ " (\~(e~ t '

\N~ "'now ~?P ) v') t\(\t, ,!:tJP ~~A'H- lOA-

S'{l Et - opp~p

IIPlug in" the values you know. Then solve for the unknown:

f5-9+<B ~'+6\(\ \1

- '0 ~\\-

~~

6 == \3+.3 -\+ l

~(j'{J ><S\(\\i

'0j~

*There is more than one right way to solve these problems!

In the last example, we could have used 73° as our angle of interest. And wecould have used the cosine (Cos e = adj/ hyp) to solve for the length.

COs~ z: 0-.. ('\J~' (OS l'~ "~l~~P ~jP

How to Solve Problems Using Angle of Elevation or Depression:

ti:<6t~o.tJJ

- Start with a diagram (right triangle)- Fill in all values that you know (into your diagram)- Decide what you are looking for (L ~ \-e.n·8 -t h?)- Choose the angle of interest and label the triangle (opp, ad], hyp)- Then choose the trig ratio to use ( remember SOH CAHTOA)

Do questions p. 132 #4- 6, 8, 10

Part 2 - Solving Right Triangles

Solving a right triangle means: find aI/lengths of \An ~ Y'\O~f\ SI0\ -e 5and find all Uw)\\nOVJo C\\1j' er- .Ex 1: Solve this triangle: What is unknown?

LPS\~e., Q~

3\d\-e. f> ~

7

Solving for the unknown angle:..-------y---y---~, Q\ \ I-'~ Clt\.O\ '-'\,p .

~'6 Q ,_..Y'~~

Solving for the unknown sides:

~~-:'....~.

.,'

~:.!::*:".:;/,,'-.>." qoO

/50):p:.i'- , ,.,"'".,"' :.,.;.,., ,.,"'.,.;"'.; :~~

L P -=- \ '6 D -- t':i 0 - Y-OLf :=. 50 -o

• Use trig ratio for the first unknown side

• For the second unknown side, use a trig ratio OR Pythagorean theorem Q

~t=\(\d. S\o.G Q 'f!..: \

Le\ ~ US(.; L Q D-S

\N ~ \\V\O\D L QJ b~f>~()\'\~j)\tlA .

(oS &-= CL('\).

\\~r

Q(\ 8\e 0-\- \ ,,-\-Lr est-

\~e W 0. {\ 1- Q:.djp opp

LOS 40 =, acb3\1-

to~ LtG '" '3,'l'\J~ :=

-----r?\-------\F\ NJ. S\ ~<- \-(~34n Pvz[ 3.1~k Lb\A\G\ use.., ,$\ (\ e .)-T(Af\e 01 ?8~tl~ ...\;htO{~tn Y\'JP

G\1.--\ '0 1.. ~ ( 1-

#t~ f'L.e. \'~ u~e. s .(\ e":: ~ S"" L\ 0 -= O\,?

h(jP 8 3,-1~\ (\L\O ~3.1 ~ oyP

~ ~. L\ \ = Q?9

Example 2: Solve this right triangle: What's unknown?

42m

You can start anywhere you like (find an angle or the missing side):

Let's start with the side (FD):

'- 2-Lt2 4- 3\1- := c.,\',\01.\ -t qlo\:: (2-

r\:).\ ~5:: A"c 2-

5~'d-- :: c.,

'I; "'~<l\nj -Mj mho (s; (\ \ LO~ or -\.on)0\(\ F =~2 ,~~.L ~ O\5~

~'P b'2· 2-o

L~=-S\(\-\ (o.5~) = 36

Find angle F:l=

Find angle D:

L D - \~D- C'\D -- 3.10

L\) 5qo

9

Solving problems with more than one right triangle:

B

A /;~c~::-:"1~~:~cFind the length BD:

Find the length AC:

~ \)\)\ ~<.e A 'J-.'~,.{c.t.s t)t \'(\-\-v -\-1) SO, v e. +V\: s -.u.sz ~ 0(\ -\\r\< v <~\'"\-\-~ A 'B D .

._ \j.S~ 0..(\!3' -{ 0..\ \ ".\Zi~.s+~ 6 e

- \4\:){\ i--(\ lAnj\<""

W C\(\~ .9\>p) nO\\JG l\jP

0\n G=2e£- S\(\2b:::~ee-h~P d-~.q

~\ (\ '26 'A 'd-.r~.q -:: DPP\ \0 ~'0 = 0 pp \

Example: Find the angle DEF: 'D

F

Notice that this is one large angle formed by the angles of each right triangle.Angle DEFis the sum of the two angles.

Call the two angles a and ~. Solve for these angles:

Cos cA =- ~J.i '-=::- <l ::: O.3by,(jP d-. 5

r). ~ Co~ ,,-\ CO, 36)

\:0<- =0 0'\ 0 \

-\-G\l\ f> =-~~0

--\-CA (\ ~:":- L\ := 0, tt 4-~

\6 ~- +~h - \ l 0 \L\-Lt )

Do questions p. 131 #1-3 ~ ~ ;;t '-\~(-\~~ \ ~ 'D Et= ::: d. -\- \3

":; (0 ~ -t ~ L\-

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