What is the csc, sec, and cot of point (3,4)?

Question

5367 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-22T20:33:22

$cosec θ = \frac{h}{o} = \frac{5}{3}$ $"cosec " theta = h/o = 5/3$

$sec θ = \frac{h}{a} = \frac{5}{4}$ $"sec " theta = h/a = 5/4$

$tan θ = \frac{a}{o} = \frac{4}{3}$ $"tan " theta = a/o = 4/3$

If you have a point $P$ $P$ at $(3, 4)$ $(3,4)$ you can draw a triangle from the origin.

A point cannot have a trig ratio, only an angle can. Assuming you mean the angle formed at $P$ $P$ . Call it $θ$ $theta$

The lengths of the sides will be as follows:

The horizontal line on the $x$ $x$ -axis will be $3$ $3$ units long.
This is the side opposite $θ$ $theta$ .

The vertical line along the line $x = 3$ $x=3$ will be $4$ $4$ units long.
This is the side adjacent to $θ$ $theta$

By Pythagoras, the hypotenuse from the origin to $P$ $P$ will be $5$ $5$ units.

NOw that we have the sides defined and we know their lengths we can determine the ratios.

$cosec θ = \frac{h}{o} = \frac{5}{3}$ $"cosec " theta = h/o = 5/3$

$sec θ = \frac{h}{a} = \frac{5}{4}$ $"sec " theta = h/a = 5/4$

$tan θ = \frac{a}{o} = \frac{4}{3}$ $"tan " theta = a/o = 4/3$