Question

What is sin θ given (–3, 4) is a point on the terminal side of θ?


Question 1 options:


3/5



-3/5



4/5



-4/5


What is cos θ given (–5, –4) is a point on the terminal side of θ?


Question 2 options:


−5√41/√41


5√41/√41


−4√41/√41


4√41/√41


The line with equation 4b + a = 0 coincides with the terminal side of an angle x in standard position in Quadrant IV.


What is sin x?


Question 3 options:


−1/4


−17/√17


√17


4√17/√17

Answer 1

Answer:

QUESTION 1-  $sin\theta =-\frac{4}{5}$

QUESTION 2-  $cos\theta =-\frac{5}{\sqrt{14} }$

QUESTION3- $sinx =-\frac{1}{\sqrt{17} }$

Step-by-step explanation:

For Question 1

From the given point(-3,4), we can deduce the following.

The point is in the 2nd quadrant.

The right-angled triangle which is formed has a hypotenuse of length 5 by Pythagoras theorem whrere B=3 and P=4.

The third quadrant are from 90° to 180° in which the angle lie.

so  $sin\theta =-\frac{P}{H}=-\frac{4}{5}$

For Question 2

From the given point(-4,-5), we can deduce the following.

The point is in the 3rd quadrant.

The right-angled triangle which is formed has a hypotenuse of length  $\sqrt{14}$ by Pythagoras theorem whrere B=-4 and P=-5.

The third quadrant are from 180° to 270° in which the angle lie.

so  $cos\theta =-\frac{P}{H} =-\frac{5}{\sqrt{14} }$

For Question 3

The line with equation 4b + a = 0 , we can deduce that the value of $b=-\frac{a}{4}$

slope of line  $tan\theta =-\frac{P}{B} =-\frac{1}{4}$

Hypotenuse of length $\sqrt{17}$ by Pythagoras theorem whrere B=4and P=1.

so  $sinx =-\frac{P}{H} =-\frac{1}{\sqrt{17} }$.