I have answered what I am permitted to: q1)

The triangle will appear in Quad III as both x and y are negative

(-3;-4).

The angle `theta` sits between the x-axis and the arm. First calculate the hypoteneuse: `(-3) ^2 + (-4)^2` `= h^2`

Therefore: h = `sqrt(25)` = 5

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I have answered what I am permitted to: q1)

The triangle will appear in Quad III as both x and y are negative

(-3;-4).

The angle `theta` sits between the x-axis and the arm. First calculate the hypoteneuse: `(-3) ^2 + (-4)^2` `= h^2`

Therefore: h = `sqrt(25)` = 5

Thus: **`sin theta = -4 /5` ** that is opposite / hypoteneuse

Now do `cos theta` and `tan theta` the same way .

For cosec sec and cot remember to invert your equation as

`cosec theta = 1/ sin theta` `sec theta = 1/ cos theta` and `cot theta = 1/ tan theta`

Thus final answer: `sin theta` = **-4 / 5**