Answer:
ii. 55 degrees.
Step-by-step explanation:
(ii)
Consider the quadrilateral ABCD.
Angle n = 180 - 96 = 84 degrees ( opposite angles of a cyclic quadrilateral are supplementary).
Consider triangle ABC: the 3 angles are supplementary, so
m < j = 180 - 84 - 41
= 55 degrees.
48.9 / 4 = 12.225
distance = 100m
time = 12.225 sec
v = d/t
The answer is 8.1799
Find the change in the y coordinates
7-5=2
Find the change in the x coordinates
1--3 or 1+3= 4
Divide the changes in the y and x coordinates
2/4= 1/2
The slope is 1/2. Hope this helps!
The midpoint of a line can be represented by the point that is in the very center of the line. A line segment such as AT also represents half of the line. The symbol of the tilde with the equal sign underneath represents congruence meaning the two segments are the same. Therefore each equation shows the same true statement in a different form
Answer:
Step-by-step explanation:
(a)
Consider the following:

Use sine rule,
![\frac{b}{a}=\frac{\sinB}{\sin A} \\\\=\frac{\sin{\frac{\pi}{3}} }{\sin{\frac{\pi}{4}}}\\\\=\frac{[\frac{\sqrt{3}}{2}]}{\frac{1}{\sqrt{2}}}\\\\=\frac{\sqrt{2}}{2}\times \frac{\sqrt{2}}{1}=\sqrt{\frac{3}{2}}](https://tex.z-dn.net/?f=%5Cfrac%7Bb%7D%7Ba%7D%3D%5Cfrac%7B%5CsinB%7D%7B%5Csin%20A%7D%0A%5C%5C%5C%5C%3D%5Cfrac%7B%5Csin%7B%5Cfrac%7B%5Cpi%7D%7B3%7D%7D%0A%7D%7B%5Csin%7B%5Cfrac%7B%5Cpi%7D%7B4%7D%7D%7D%5C%5C%5C%5C%3D%5Cfrac%7B%5B%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%5D%7D%7B%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D%7D%5C%5C%5C%5C%3D%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%5Ctimes%20%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B1%7D%3D%5Csqrt%7B%5Cfrac%7B3%7D%7B2%7D%7D)
Again consider,
![\frac{b}{a}=\frac{\sin{B}}{\sin{A}} \\\\\sin{B}=\frac{b}{a}\times \sin{A}\\\\\sin{B}=\sqrt{\frac{3}{2}}\sin {A}\\\\B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]](https://tex.z-dn.net/?f=%5Cfrac%7Bb%7D%7Ba%7D%3D%5Cfrac%7B%5Csin%7BB%7D%7D%7B%5Csin%7BA%7D%7D%0A%5C%5C%5C%5C%5Csin%7BB%7D%3D%5Cfrac%7Bb%7D%7Ba%7D%5Ctimes%20%5Csin%7BA%7D%5C%5C%5C%5C%5Csin%7BB%7D%3D%5Csqrt%7B%5Cfrac%7B3%7D%7B2%7D%7D%5Csin%20%7BA%7D%5C%5C%5C%5CB%3D%5Csin%5E%7B-1%7D%5B%5Csqrt%7B%5Cfrac%7B3%7D%7B2%7D%7D%5Csin%7BA%7D%5D)
Thus, the angle B is function of A is, ![B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]](https://tex.z-dn.net/?f=B%3D%5Csin%5E%7B-1%7D%5B%5Csqrt%7B%5Cfrac%7B3%7D%7B2%7D%7D%5Csin%7BA%7D%5D)
Now find 
Differentiate implicitly the function
with respect to A to get,

b)
When
, the value of
is,

c)
In general, the linear approximation at x= a is,

Here the function ![f(A)=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]](https://tex.z-dn.net/?f=f%28A%29%3DB%3D%5Csin%5E%7B-1%7D%5B%5Csqrt%7B%5Cfrac%7B3%7D%7B2%7D%7D%5Csin%7BA%7D%5D)
At 
![f(\frac{\pi}{4})=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{\frac{\pi}{4}}]\\\\=\sin^{-1}[\sqrt{\frac{3}{2}}.\frac{1}{\sqrt{2}}]\\\\\=\sin^{-1}(\frac{\sqrt{2}}{2})\\\\=\frac{\pi}{3}](https://tex.z-dn.net/?f=f%28%5Cfrac%7B%5Cpi%7D%7B4%7D%29%3DB%3D%5Csin%5E%7B-1%7D%5B%5Csqrt%7B%5Cfrac%7B3%7D%7B2%7D%7D%5Csin%7B%5Cfrac%7B%5Cpi%7D%7B4%7D%7D%5D%5C%5C%5C%5C%3D%5Csin%5E%7B-1%7D%5B%5Csqrt%7B%5Cfrac%7B3%7D%7B2%7D%7D.%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D%5D%5C%5C%5C%5C%5C%3D%5Csin%5E%7B-1%7D%28%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%29%5C%5C%5C%5C%3D%5Cfrac%7B%5Cpi%7D%7B3%7D)
And,
from part b
Therefore, the linear approximation at
is,
![f(x)=f'(A).(x-A)+f(A)\\\\=f'(\frac{\pi}{4}).(x-\frac{\pi}{4})+f(\frac{\pi}{4})\\\\=\sqrt{3}.[x-\frac{\pi}{4}]+\frac{\pi}{3}](https://tex.z-dn.net/?f=f%28x%29%3Df%27%28A%29.%28x-A%29%2Bf%28A%29%5C%5C%5C%5C%3Df%27%28%5Cfrac%7B%5Cpi%7D%7B4%7D%29.%28x-%5Cfrac%7B%5Cpi%7D%7B4%7D%29%2Bf%28%5Cfrac%7B%5Cpi%7D%7B4%7D%29%5C%5C%5C%5C%3D%5Csqrt%7B3%7D.%5Bx-%5Cfrac%7B%5Cpi%7D%7B4%7D%5D%2B%5Cfrac%7B%5Cpi%7D%7B3%7D)
d)
Use part (c), when
, B is approximately,
![B=f(46°)=\sqrt{3}[46°-\frac{\pi}{4}]+\frac{\pi}{3}\\\\=\sqrt{3}(1°)+\frac{\pi}{3}\\\\=61.732°](https://tex.z-dn.net/?f=B%3Df%2846%C2%B0%29%3D%5Csqrt%7B3%7D%5B46%C2%B0-%5Cfrac%7B%5Cpi%7D%7B4%7D%5D%2B%5Cfrac%7B%5Cpi%7D%7B3%7D%5C%5C%5C%5C%3D%5Csqrt%7B3%7D%281%C2%B0%29%2B%5Cfrac%7B%5Cpi%7D%7B3%7D%5C%5C%5C%5C%3D61.732%C2%B0)