Let \(\displaystyle t\) stand for the length of \(\displaystyle \overline{AX}\); then the length of \(\displaystyle \overline{CX}\) is twice this, or \(\displaystyle 2t\). The figure referenced is below:

If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,

\(\displaystyle AX \cdot BX = CX \cdot DX\)

Substituting the appropriate quantities, then solving for \(\displaystyle t\):

\(\displaystyle t \cdot 20 = 2t \cdot 10\)

\(\displaystyle 20 t = 20t\)

This statement is identically true. Therefore, without further information, we cannot determine the value of \(\displaystyle t\) - the length of \(\displaystyle \overline{AX}\).

Let \(\displaystyle t\) stand for the length of \(\displaystyle \overline{AX}\); then the length of  \(\displaystyle \overline{BX}\) is \(\displaystyle t+12\). The figure referenced is below:

If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,

\(\displaystyle AX \cdot BX = CX \cdot DX\)

Substituting the appropriate quantities, then solving for \(\displaystyle t\):

\(\displaystyle t \cdot (t+12) = 8 \cdot 10\)

\(\displaystyle t \cdot t+t \cdot 12 = 8 0\)

\(\displaystyle t ^{2}+12 t = 8 0\)

This quadratic equation can be solved by completing the square; since the coefficient of \(\displaystyle t\) is 12, the square can be completed by adding

\(\displaystyle \left (\frac{12}{2} \right ) ^{2} = 6 ^{2} = 36\)

to both sides:

\(\displaystyle t ^{2}+12 t + 36 = 8 0+ 36\)

Restate the trinomial as the square of a binomial:

\(\displaystyle (t+6 )^{2 } = 116\)

Take the square root of both sides:

\(\displaystyle t+6 = \pm \sqrt{116}\)

\(\displaystyle t+6 \approx - 10.8\) or  \(\displaystyle t+6 \approx 10.8\)

Either

\(\displaystyle t+6 \approx - 10.8\), 

in which case

\(\displaystyle t+6 - 6 \approx - 10.8 - 6\)

\(\displaystyle t \approx -16.8\),

or 

\(\displaystyle t+6 \approx 10.8\)

in which case

\(\displaystyle t+6 - 6 \approx 10.8 - 6\)

\(\displaystyle t \approx 4.8\),

Since \(\displaystyle t\) is a length, we throw out the negative value; it follows that \(\displaystyle t \approx 4.8\), the correct length of \(\displaystyle \overline{AX}\).

In a circle, a diameter perpendicular to a chord bisects the chord. This makes \(\displaystyle X\) the midpoint of \(\displaystyle \overline{CD}\); consequently, \(\displaystyle CX =DX = \frac{1}{2} CD = \frac{1}{2} \cdot 20 = 10\).

The figure referenced is below:

If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,

\(\displaystyle AX \cdot BX = CX \cdot DX\)

Setting  \(\displaystyle AX= 8, CX =DX = 10\), and solving for \(\displaystyle BX\):

\(\displaystyle 8 \cdot BX = 10 \cdot 10\)

\(\displaystyle 8 \cdot BX = 100\)

\(\displaystyle 8 \cdot BX \div 8 = 100 \div 8\)

\(\displaystyle BX = 12.5\)

\(\displaystyle AB = AX + BX = 8 + 12.5 = 20.5\),

the correct length.

Let \(\displaystyle t = AX\), in which case \(\displaystyle CX = t+ 6\); the figure referenced is below (not drawn to scale). 

If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,

\(\displaystyle AX \cdot BX = CX \cdot DX\)

Setting  \(\displaystyle AX= t, CX= t+ 6, BX = 12, DX = 10\), and solving for \(\displaystyle t\):

\(\displaystyle AX \cdot BX = CX \cdot DX\)

\(\displaystyle t \cdot 12 = (t+6) \cdot 10\)

\(\displaystyle 12t = 10t + 60\)

\(\displaystyle 12t- 10t = 10t + 60 - 10t\)

\(\displaystyle 2t = 60\)

\(\displaystyle \frac{2t}{2} = \frac{60}{2}\)

\(\displaystyle t = 30\), 

which is the length of \(\displaystyle \overline{AX}\).

A diameter of a circle perpendicular to a chord bisects the chord. Therefore, the point of intersection \(\displaystyle X\) is the midpoint of \(\displaystyle \overline{CD}\), and

\(\displaystyle CX = DX = \frac{1}{2} CD\). 

Let \(\displaystyle t\) stand for the common length of \(\displaystyle \overline{CX}\) and \(\displaystyle \overline{DX}\),

The figure referenced is below.

If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,

\(\displaystyle CX \cdot DX = AX \cdot BX\)

Set \(\displaystyle AX = 12\) and \(\displaystyle BX = 20\), and \(\displaystyle CX = DX = t\); substitute and solve for \(\displaystyle t\):

\(\displaystyle t \cdot t = 12 \cdot 20\)

\(\displaystyle t^{2 } = 240\)

\(\displaystyle t = \sqrt{240} \approx 15.5\)

This is the length of \(\displaystyle \overline{CX}\); the length of \(\displaystyle \overline{CD}\) is twice this, so

\(\displaystyle CD = 2 \cdot t \approx 2 \cdot 15.5 = 31.0\)

The measure of the angle formed by the two secants to the circle from a point outside the circle is equal to half the difference of the two arcs they intercept; that is,

\(\displaystyle m \angle BND = \frac{1}{2} ( \overarc{BD} - \overarc{AC})\)

The ratio of the degree measure of \(\displaystyle \overarc{BD}\) to that of \(\displaystyle \overarc{AC}\) is that of their lengths, which is 7 to 2. Therefore,

\(\displaystyle \frac{m \overarc{BD} }{m \overarc{AC} } = \frac{7}{2}\)

Letting \(\displaystyle t = m \overarc{AC}\):

\(\displaystyle \frac{m \overarc{BD} }{t } = \frac{7}{2}\)

\(\displaystyle \frac{m \overarc{BD} }{t } \cdot t = \frac{7}{2} \cdot t\)

\(\displaystyle m \overarc{BD} = \frac{7}{2} t\)

Therefore, in terms of \(\displaystyle t\):

\(\displaystyle m \angle BND = \frac{1}{2} \left ( \frac{7}{2} t - t \right ) = \frac{1}{2} \left ( \frac{5}{2} t \right ) =\frac{5}{4} t\)

Without further information, however, we cannot determine the value of \(\displaystyle t\) or that of \(\displaystyle m \angle BND\). Therefore, the given information is insufficient.

At four o’clock the minute hand is on the 12 and the hour hand is on the 4.  The angle formed is 4/12 of the total number of degrees in a circle, 360.

4/12 * 360 = 120 degrees

A clock is a circle, and a circle always contains 360 degrees. Since there are 60 minutes on a clock, each minute mark is 6 degrees.

\(\displaystyle \frac{360^o\ \text{total}}{60\ \text{minutes total}}=6\ \text{degrees per minute}\)

The minute hand on the clock will point at 15 minutes, allowing us to calculate it's position on the circle.

\(\displaystyle (15\ min)(6)=90^o\)

Since there are 12 hours on the clock, each hour mark is 30 degrees.

\(\displaystyle \frac{360^o\ \text{total}}{12\ \text{hours total}}=30\ \text{degrees per hour}\)

We can calculate where the hour hand will be at 8:00.

\(\displaystyle (8\ hr)(30)=240^o\)

However, the hour hand will actually be between the 8 and the 9, since we are looking at 8:15 rather than an absolute hour mark. 15 minutes is equal to one-fourth of an hour. Use the same equation to find the additional position of the hour hand.

\(\displaystyle 240^o+(\frac{1}{4}\ hr)(30)\)

\(\displaystyle 240^o+7.5^o\)

\(\displaystyle 247.5^o\)

We are looking for the angle between the two hands of the clock. The will be equal to the difference between the two angle measures.

\(\displaystyle \angle=247.5^o-90^o=157.5^o\)

A analog clock is divided up into 12 sectors, based on the numbers 1–12. One sector represents 30 degrees (360/12 = 30). If the hour hand is directly on the 10, and the minute hand is on the 2, that means there are 4 sectors of 30 degrees between then, thus they are 120 degrees apart (30 * 4 = 120).

At \(\displaystyle 5:00\), the hour hand is on the \(\displaystyle 5\) and the minute hand is at the \(\displaystyle 12\). There are \(\displaystyle 12\) spaces on a clock, and these hands are separated by \(\displaystyle 5\) spaces.

Thus, the angle between them is \(\displaystyle \frac{5}{12}\) the degrees of the entire clcok, which is \(\displaystyle 360^\circ\).

Therefore, we multiply these to get our answer. 

\(\displaystyle \frac{5}{12}\times \frac{360}{1}\) 

We can cancel out as we multiply to get: 

\(\displaystyle \frac{5}{1}\times \frac{30}{1}=150^\circ\)