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Tasks like this one are difficult for me. I have been looking at this task alone for a few hours now. enter image description here

g) The circle has the center at O. AB cuts of a 60 degrees section, while DE cuts of a 20 degree section.

The rest is just like this; bestem = find. Vis at = show that.

The only thing I see is that $\angle AOB$ is $60^\circ$ And from that there is a rule that should make $\angle ADB = 30^\circ$ But I'm not convinced that is correct. I can't find a word for that rule in english. Wiki gets complicated really fast about even simple stuff.

share|cite|improve this question – y zh May 28 '12 at 16:55
There is such a rule. If you have a chord $AB$, and a point like $D$, then the angle subtended by $AB$ at the centre of the circle (that is, $\angle AOB$) is twice $\angle ADB$. Since $\angle AOB=60^\circ$, you are right in asserting that $\angle ADB=30^\circ$. – André Nicolas May 28 '12 at 16:59
I think the use of words like "ass" should be avoided here. – smanoos May 28 '12 at 17:01
Similarly, $\angle DAE=10^\circ$. That should give you plenty of information to chase down the rest. – André Nicolas May 28 '12 at 17:01
I forget how dandy the U.S is compared to the UK. My bad. Thanks for the help @AndréNicolas – Algific May 28 '12 at 17:06

The rule that you describe, that says the measure of an inscribed angle (in this case, $\angle ADB$) is half the measure of the central angle ($\angle AOB$) that subtends the same arc ($\stackrel\frown{AB}$) is often called the Inscribed Angle Theorem.

As you said, that gets you (1); using the rule with the other given arc will get you (2). As to part (3), there is a theorem that gives the result directly (hidden in a "hint" box at the bottom of this answer), but your translation of "Show that..." suggests to me that you aren't supposed to use the theorem.

Hint as to how to do part (3):

Consider $\triangle BCD$, particularly the measures of its angles.

The theorem (for any picture such as the one you posted, regardless of the given angle measures):

$$m\angle ACB=\frac{m\stackrel\frown{AB}-m\stackrel\frown{DE}}{2}.$$

(Also, $$m\angle ASB=\frac{m\stackrel\frown{AB}+m\stackrel\frown{DE}}{2}$$.)

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