Question

Super stuck right now... Help will be greatly appreciated :)
2 part question attached below .

Answer 1

could you find L ?

Answer 2

Noppee. I don't know how to show what L equals to.

Answer 3

Consider triangle BAR,
What will be sin theta, in terms of BR?

Answer 4

4/BR ?

Answer 5

Note that L= PR
and yes,
thats correct,
so we get
BR = 4/sin theta


In triangle PAR,
what will be cos theta?

Answer 6

POR, i meant**

Answer 7

oh, and I think we will need AR,
in terms of cos theta from triangle BAR too

Answer 8

okay.. Cos theta in triangle BAR would be AR/BR
And then cos theta in triangle POR would be (2 + AR)/PR

Answer 9

lets gather what all information we have till now,

\(BR = \dfrac{4}{\sin \theta } \\ AR = BR \cos \theta \\ PR = (2+AR)\cos \theta \)

Answer 10

got those?

now you'll just have to do all the plugging in business...
like
plug in BR = 4/ sin theta in
AR =BR cos theta, equation.

Answer 11

yep i got them.
oh okaay. Hang on!

Answer 12

i made a typo
\(2+AR = PR \cos \theta \\ \Large PR =\dfrac{2+AR}{\cos \theta } \)

Answer 13

yeah xD i was wondering because i subbed it in and i was like...... hahha.
okayyy!

Answer 14

so ladder length L =
\(\Large L = PR = \dfrac{2}{\cos \theta }+\dfrac{AR}{\cos \theta }\)

Answer 15

just plug in for AR!

Answer 16

yes! I got it:) Thanks!.

Answer 17

how about the other parT?
dL/dtheta ?

Answer 18

Stuck on that too LOL

Answer 19

have you learnt derivatives?
u/v rule ?