Can anyone explain these to me plz don’t just give the answers

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[dim dune]

@lavish tendon try drawing the triangle the fallen branch makes with the tree and label the angle and distance given

[lavish tendon]

i tried that

the issue is

i found both legs of the right triangle to be equal

because

tan45=1

and tan 30 doesnt include any variables

[dim dune]

This is what you have right

And we need to find the distance RG

Rock ground

And yes, you can use trig or geometry to find it

Specifically cosine would help because it's formula includes the adjacent side (which is given to us as 12.3) and the hypotenuse, which we need to find

[left pine]

yo can u help me after ur done

[dim dune]

Another way would be to mirror the triangle at angle 30° to get an equilateral triangle. Then you can find the side from the formula for height

[dim dune]

Maybe, make your own thread

[left pine]

ok

[lavish tendon]

not exactly

let me show u

[dim dune]

Okay and where is the stone

[lavish tendon]

at the left side of the line x+12.3

[dim dune]

I understood it as being on top of the tree, which would be y

And z would be equal to y

[lavish tendon]

in my diagram y is meant to be the height of the tree

[lavish tendon]

so what do i do based on my diagram

[dim dune]

Im guessing you can find z

y=x+12.3

[dim dune]

The confusion for me here is the text. Its not clear to me whether the stone is on top of the tree, and whether the branch falls behind that 45° angle

[lavish tendon]

i think

the stone is on the floor

[dim dune]

So right side of the line

[lavish tendon]

can u try the other question

[lavish tendon]

yes

[lavish tendon]

look in my diagram how would u find x

[dim dune]

If the stone is on the floor where the tree is, then the distance to the branch where it touches the ground is just 12.3

According to your image

[lavish tendon]

wait howww

[dim dune]

Is the rock at the 90° angle?

[lavish tendon]

no

[lavish tendon]

no

[dim dune]

for the first one I don't know, still feels like you're assuming the rock is on the ground when it says the stone is on top of the tree. might be a language issue on my end.
for the second, turn cosec and sec into 1/sin and 1/cos and square them both

@lavish tendon

so $(\cos{A} - \csc{A})^2 + (\sin{A} - \sec{A})^2 = \left(\cos{A} - \frac{1}{\sin{A}}\right)^2 + \left(\sin{A} - \frac{1}{\cos{A}}\right)^2$

[lavish tendon]

i tried this

but

after u simplify

it doesnt give u smth thats part of the options

[limpid spire]

Ephesians 2:8-9

[dim dune]

ok lets square them

$\cos^2{A} - \frac{2\cos{A}}{\sin{A}} + \frac{1}{\sin^2{A}} + \sin^2{A} - \frac{2\sin{A}}{\cos{A}} + \frac{1}{\cos^2{A}}$

is that right

[limpid spire]

Ephesians 2:8-9

[lavish tendon]

js change the cos alpha to cos a

ye

[dim dune]

yeah got it

[lavish tendon]

thats what i got

[dim dune]

ok, and you didn't get further than that?

[lavish tendon]

then sin^2(x)+cos^2(x)=1

[dim dune]

true

[lavish tendon]

the middle terms

i wrote as

[dim dune]

$1 - \frac{2\cos{A}}{\sin{A}} + \frac{1}{\sin^2{A}} - \frac{2\sin{A}}{\cos{A}} + \frac{1}{\cos^2{A}}$

[limpid spire]

Ephesians 2:8-9

[lavish tendon]

-2(cotx+tanx)

for the middle terms

i think that might have been a mistake

but im not sure

ik the maths is right

but

it might not be the approach ur meant to take

[dim dune]

$1 - 2\left(\frac{\cos{A}}{\sin{A}}+\frac{\sin{A}}{\cos{A}}\right) + \frac{1}{\sin^2{A}} + \frac{1}{\cos^2{A}}$

whoops

[lavish tendon]

yep

[dim dune]

try to find the common denominator for those fractions

[lavish tendon]

U forgot then add 1/cos^2(x)

[dim dune]

yeah just saw

[limpid spire]

Ephesians 2:8-9

[dim dune]

anyways, try and find the common denominator

and on top you will get 1

[lavish tendon]

would it be cos squared sin squared

[dim dune]

for the second part yeah

[lavish tendon]

sinacosa

oh

ye i got it

[dim dune]

great

[lavish tendon]

so ur left with

-2/cosasina

[lavish tendon]

what do we do next

[dim dune]

$1 - 2\left(\frac{1}{\sin{A}\cos{A}}\right) + \frac{1}{\sin^2{A}\cos^2{A}}$

[limpid spire]

Ephesians 2:8-9

[dim dune]

that's what you're left with right

turn the 1/sinAcosA into sec and cosec

same with the square

and you should be able to turn it into one of the forms from the answers

[lavish tendon]

So would it be

[lavish tendon]

1-2seccosec+sec^2tan^2

srry

[dim dune]

not tan yeah

[lavish tendon]

cosec

[dim dune]

correct

$1 - 2\sec{A}\csc{A} + \sec^2{A}\csc^2{A}$

[limpid spire]

Ephesians 2:8-9

[lavish tendon]

OHH

I GET IT NOW

[dim dune]

$1 - 2\sec{A}\csc{A} + (\sec{A}\csc{A})^2$ maybe that makes it easier to see what it is

[limpid spire]

Ephesians 2:8-9

[lavish tendon]

thank you sm

[dim dune]

np, sorry about the first problem though

[lavish tendon]

all good

[lavish tendon]

thank you so much

[ripe skiff]

@lavish tendon has given 1 rep to @dim dune

[dim dune]

yw anytime

feel free to repost the first problem, I don't think anyone will look in here

[lavish tendon]

i will

[dim dune]

+close