(a) In the diagram. \(\over{Rs}\) and \(\over{RT}\) are tangent to the circle with centre O, < TUS = 68\(^o\), < SRT = x and < UTO = y. Find the value of x....

Answer Details

(a)

In this problem, we have a circle with center O and two tangent lines, Rs and RT. The angles between these lines and the line connecting the center of the circle to the point of tangency are given as TUS = 68°, UTO = y, and SRT = x.

Since Rs and RT are tangent to the circle, the angle between the line connecting the center of the circle to the point of tangency and the line connecting the center of the circle to the point of tangency on the other line is equal to 90°.

Therefore, we have the following two equations:

UTO + TUS = 90° (angle sum in a semicircle)

SRT + TUS = 90° (angle sum in a semicircle)

Substituting the given values, we get:

y + 68° = 90°

and

x + 68° = 90°

Solving for x and y, we get:

x = 90° - 68° = 22°

y = 90° - 68° = 22°

Therefore, the value of x is 22°.

(b)

Let's call the capacity of tank A as "A" and the capacity of tank B as "B".

Given that:

A = B + 600 (tank A holds 600 litres more than tank B)

And, after pumping out 100 litres from each tank:

A - 100 = 3 (B - 100)

Substituting the first equation into the second equation, we get:

B + 500 = 3 (B - 100)

Expanding the second term on the right side, we get:

B + 500 = 3B - 300

Solving for B, we get:

2B = 800

B = 400

Therefore, the capacity of tank B is 400 litres.

And, using the first equation, we get:

A = B + 600 = 400 + 600 = 1000

Therefore, the capacity of tank A is 1000 litres.