Consider the absolute value, because we only worry about the quadrant later.
![\text{Consider: } sin(A) = |-\frac{\sqrt{3}}{4}|](https://tex.z-dn.net/?f=%5Ctext%7BConsider%3A%20%7D%20sin%28A%29%20%3D%20%7C-%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B4%7D%7C)
![sin(A) = \frac{\sqrt{3}}{4}](https://tex.z-dn.net/?f=sin%28A%29%20%3D%20%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B4%7D)
Thus, we know that the hypotenuse has a length of 4 units, and the side opposite the angle, A is √3, because this is the nature of the sine function in relation to its triangular component.
The missing side can be found using Pythagoras' Theorem:
4² - (√3)² = x²
16 - 3 = x²
13 = x²
x = √13
![\therefore tan(A) = \pm \sqrt{\frac{3}{13}}](https://tex.z-dn.net/?f=%5Ctherefore%20tan%28A%29%20%3D%20%5Cpm%20%5Csqrt%7B%5Cfrac%7B3%7D%7B13%7D%7D)
Since angle A is in the third quadrant, the tangent function will produce a positive angle.
![tan(A) = \sqrt{\frac{3}{13}}](https://tex.z-dn.net/?f=tan%28A%29%20%3D%20%5Csqrt%7B%5Cfrac%7B3%7D%7B13%7D%7D)
![\text{Rationalise the denominator: } \frac{\sqrt{3}}{\sqrt{13}} \cdot \frac{\sqrt{13}}{\sqrt{13}}](https://tex.z-dn.net/?f=%5Ctext%7BRationalise%20the%20denominator%3A%20%7D%20%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B%5Csqrt%7B13%7D%7D%20%5Ccdot%20%5Cfrac%7B%5Csqrt%7B13%7D%7D%7B%5Csqrt%7B13%7D%7D)
Answer:
It should be C. 620
Step-by-step explanation:
Im not sure what slope means but if it means gradient then the gradient is 3x.
Answer:
x > −4
Step-by-step explanation:
Step 1: Add 5 to both sides.
6x-5+5 > -29+5
6x>-24
Step 2: Divide both sides by 6
6x/6 > -24/6
x > -4