I don’t know but I’m going to say that the answer is 48? I don’t know. If I’m wrong please help me understand this. Lol this is the same thing I’m doing right now and the question I was actually about to ask
The answer is x<0 and y<0
When you reflect a line segment, it would create a mirror-image of itself about the specified axis. Thus, if you reflect it over the x-axis, it will now be situated at the 4th quadrant, because it is what's opposite to the 1st quadrant with respect to the x-axis.
Tan (Ф/2)=⁺₋√[(1-cosФ)/(1+cosФ)]
if π<Ф<3π/2;
then, Where is Ф/2??
π/2<Ф/2<3π/4; therefore Ф/2 is in the second quadrant; then tan (Ф/2) will have a negative value.
tan(Ф/2)=-√[(1-cosФ)/(1+cosФ)]
Now, we have to find the value of cos Ф.
tan (Ф)=4/3
1+tan²Ф=sec²Ф
1+(4/3)²=sec²Ф
sec²Ф=1+16/9
sec²Ф=(9+16)/9
sec²Ф=25/9
sec Ф=-√(25/9) (sec²Ф will have a negative value, because Ф is in the sec Ф=-5/3 third quadrant).
cos Ф=1/sec Ф
cos Ф=1/(-5/3)
cos Ф=-3/5
Therefore:
tan(Ф/2)=-√[(1-cosФ)/(1+cosФ)]
tan(Ф/2)=-√[(1+3/5)/(1-3/5)]
tan(Ф/2)=-√[(8/5)/(2/5)]
tan(Ф/2)=-√4
tan(Ф/2)=-2
Answer: tan (Ф/2)=-2; when tan (Ф)=4/3
63.5 divided by 5 is 12.7.