In the diagram, \(\overline{RT}\) and \(\overline{RT}\) are tangent to the circle with centre O. < TUS = 68 °, < SRT = x, and < UTO = y. Find the value of x.
(b) Two tanks A and B am filled to capacity with diesel. Tank A holds 600 litres of diesel more than tank B. If 100 litres of diesel was pumped out of each tank, tank A would then contain 3 times as much diesel as tank B. Find the capacity of each tank.
(a) Since \(\overline{RT}\) and \(\overline{RU}\) are tangent to the circle, \(\angle TRU\) is equal to the angle between the radii drawn to the points of tangency. Therefore, \(\angle TRU = \angle TRO = 90^{\circ}\).
Also, since \(\overline{RT}\) is tangent to the circle, \(\angle SRT\) is equal to the angle between the tangent and the chord \(\overline{ST}\). Therefore, \(\angle SRT = \angle OTU\).
Finally, since the sum of the angles in a triangle is 180^{\circ}, we have:
\[\angle TUS + \angle SRT + \angle UTO = 180^{\circ}\]
\[68^{\circ} + x + y = 180^{\circ}\]
\[x = 112^{\circ} - y\]
Therefore, the value of x is 112^{\circ} - y.
(b) Let the capacity of tank B be x litres. Then, the capacity of tank A is x+600 litres.
After 100 litres of diesel is pumped out of each tank, tank B has (x-100) litres of diesel and tank A has (x+500) litres of diesel. Since tank A has three times as much diesel as tank B, we can set up the following equation:
\[(x+500) = 3(x-100)\]
Expanding the right-hand side gives:
\[x + 500 = 3x - 300\]
Simplifying and solving for x gives:
\[x = 400\]
Therefore, the capacity of tank B is 400 litres and the capacity of tank A is x+600 = 1000 litres.
