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)
81.
Add them all and then divide by the number of days (by 5)
Answer: The answer is (B) ∠SYD.
Step-by-step explanation: As mentioned in the question, two parallel lines PQ and RS are drawn in the attached figure. The transversal CD cut the lines PQ and RS at the points X and Y respectively.
We are given four angles, out of which one should be chosen which is congruent to ∠CXP.
The angles lying on opposite sides of the transversal and outside the two parallel lines are called alternate exterior angles.
For example, in the figure attached, ∠CXP, ∠SYD and ∠CXQ, ∠RYD are pairs of alternate exterior angles.
Now, the theorem of alternate exterior angles states that if the two lines are parallel having a transversal, then alternate exterior angles are congruent to each other.
Thus, we have
∠CXP ≅ ∠SYD.
So, option (B) is correct.
Answer:
1 1/3 hours
Step-by-step explanation:
Pipe 1 alone:
fills the pool in 2 hours
in 1 hour, it fills 1/2 of the pool
Pipe 2 alone:
fills the pool in 4 hours
in 1 hour, it fills 1/4 of the pool
Pipe 1 and Pipe 2 working together:
fill the pool in x hours
in 1 hour, they fill 1/x of the pool
Working together, in 1 hour the two pipes fill 1/2 + 1/4 of the pool.
Working together, in 1 hour the two pipes fill 1/x of the pool.
Therefore,
1/2 + 1/4 = 1/x
2/4 + 1/4 = 1/x
3/4 = 1/x
x = 4/3
x = 1 1/3
Answer: 1 1/3 hours