First note that a clock is a circle made of 360 degrees, and that each number represents an angle and the separation between them is 360/12 = 30. And at 2:00, the minute hand is on the 12 and the hour hand is on the 2.  The correct answer is 2 * 30 = 60 degrees.

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).

First, we must determine where the hands are located at this time. The minute hand will be exactly on the three. The hour hand will be a little bit (15 mins) passed the seven. A circle has \(\displaystyle 360^{\circ}\) in it, so because there are \(\displaystyle 12\) numbers on a clock's face, each number is separated by

\(\displaystyle 360^{\circ}\div12^{\circ}=30^{\circ}\)

The angle between the \(\displaystyle 3\) and the \(\displaystyle 7\) is equal to \(\displaystyle 4\times30^{\circ}=120^{\circ}\) 

Finally, we must figure out exactly how far past the \(\displaystyle 7\) our hour hand is. If the hour hand moves \(\displaystyle 30^{\circ}\) in \(\displaystyle 60\textup{ minutes}\), that means that it must move a half of a degree every minute. If we are \(\displaystyle 15\textup{ minutes}\) past \(\displaystyle 7\textup{:}00\), we can do \(\displaystyle \left ( \frac{1}{2} \right )15=7.5\)

Therefore, our hour hand has moved an extra \(\displaystyle 7.5^{\circ}\) due to the \(\displaystyle 15\textup{ minutes}\) that have elapsed.

Our angle is equal to 

\(\displaystyle 120^{\circ}+7.5^{\circ}=127.5^{\circ}\).

The entire clock measures 360°. As the clock is divided into 12 sections, the distance between each number is equivalent to 30° (360/12). The distance between the 2 and the 3 on the clock is 30°.  One has to account, however, for the 10 minutes that have passed. 10 minutes is 1/6 of an hour so the hour hand has also moved 1/6 of the distance between the 3 and the 4, which adds 5° (1/6 of 30°). The total measure of the angle, therefore, is 35°.

To solve a chord problem, draw right triangles using the chord, the radii, and a line connecting the center of the circle to the chord at a right angle. 

Now, the chord is split into two equal pieces, and angle AOB is bisected. Instead of one 120 degree angle, you now have two 30-60-90 triangles. 30-60-90 triangles are characterized by having sides in the following ratio:

So, to find the length of the chord, first find the length of each half. Because the triangles in your circle are similar to the 30-60-90 triangle above, you can set up a proportion. The hypotenuse of our triangle is 6 (the radius of the circle) so it is set over 2 (the hypotenuse of our model 30-60-90 triangle). Half of the chord of the circle is the leg of the triangle that is across from the 60 degree angle (120/2), so it corresponds to the \(\displaystyle \sqrt{3}\) side of the model triangle. 

\(\displaystyle \frac{x}{6}=\frac{\sqrt{3}}{2}\)

Therefore, \(\displaystyle x=3\sqrt{3}\)

Because x is equal to half of the chord, the answer is \(\displaystyle 6\sqrt{3}\).

The formula for the area of a circle is \(\displaystyle A=\pi r^2\), and the formula for circumference is \(\displaystyle C=2\pi r\). If we solve for C in terms of r, we get
\(\displaystyle r=C/2\pi\).

We can then substitute this value of r into the formula for the area:

\(\displaystyle A=\pi r^2\)

\(\displaystyle =\pi (C/2\pi )^2\)

\(\displaystyle =C^2\pi /4\pi ^2\)

\(\displaystyle =C^2/4\pi\)

Set the area of the circle equal to four times the circumference πr² = 4(2πr). 

Cross out both π symbols and one r on each side leaves you with r = 4(2) so r = 8 and therefore d = 16.

The perimeter of a circle = 2 πr = πd

Therefore d = 36