please help gr 11 trig

[upper yew]

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[deep spindle]

so we're working with this picture yeah

[cosmic lotus]

Yes

[deep spindle]

what's the initial arm here

[cosmic lotus]

what do you mean?

[deep spindle]

(never heard that term before honestly)

[cosmic lotus]

like

[deep spindle]

here like this

where are we measuring theta from

[cosmic lotus]

q3

[deep spindle]

so then do we want this angle?

[cosmic lotus]

yes

[deep spindle]

should just be arctan(1/√3)

[cosmic lotus]

ill show u what i did

-i got tan (1/root 3) which is 7pi/6

[deep spindle]

so 30°

[grim zephyr]

(\frac{dy}{dx}=\frac{1}{\sqrt{3}}=tan(\theta))

[cosmic violet]

Henry Rollins

[cosmic lotus]

i did not do calc yet

ir derivatives

[deep spindle]

how did you get 7π/6
it's just π/6

which in degrees is 30°

[cosmic lotus]

yeah but Q3

[deep spindle]

i asked you where we're measuring from

[cosmic lotus]

idk what the question is asking

[deep spindle]

are we measuring this then?

[deep spindle]

neither do i

weird wording

[cosmic lotus]

the question is 4 marks

[grim zephyr]

Graphing calculator?

[cosmic lotus]

so i am assuming

they mean the full arm

This is what I did

The angle stays the same regardless of the C value

But not sure if it’s fully correct @deep spindle

[grim zephyr]

But the angle is 30° right?

[cosmic lotus]

It’s 210

[grim zephyr]

How?

[cosmic lotus]

It doesn’t ask for the related acure angle

Q3

[grim zephyr]

Oh OK

[deep spindle]

well then its 210°