Gleichschenkliges Dreieck
About points...
We associate a certain number of points with each exercise.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
About difficulty...
We associate a certain difficulty with each exercise.
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
Question
Solution
Short
Video
\(\LaTeX\)
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Exercise:
minipage.textwidth Betrachte das nebenstehe gleichschenklige Dreieck mit der Basis b den Schenkeln a und den Winkeln alpha beta. abclist abc Bestimme die Winkel alpha und beta wenn bcm und acm ist. abc Bestimme die Länge a wenn alpha^circ und b.cm ist. abclist Hinweis: Zerlege das Dreieck in zwei rechtwinklige Dreiecke. minipage hfill minipage.textwidth raggedleft GleichschenkligesDreieck. minipage
Solution:
abclist abc Wir zeichnen die Höhe h ein wodurch sowohl der Winkel alpha als auch die Basis b halbiert werden. Damit ist a die Hypotenuse und frac b eine Gegenkathete zum Winkel fracalpha respektive eine Ankathete zum Winkel beta. Damit folgt: sintfracalpha fracfrac ba frac &&|AN tfracalpha ^circ &&| alpha ^circ beta ^circ - fracalpha ^circ. abc Die Länge a ist a fracfrac bsintfracalpha frac.cmsin^circ .cm. abclist
minipage.textwidth Betrachte das nebenstehe gleichschenklige Dreieck mit der Basis b den Schenkeln a und den Winkeln alpha beta. abclist abc Bestimme die Winkel alpha und beta wenn bcm und acm ist. abc Bestimme die Länge a wenn alpha^circ und b.cm ist. abclist Hinweis: Zerlege das Dreieck in zwei rechtwinklige Dreiecke. minipage hfill minipage.textwidth raggedleft GleichschenkligesDreieck. minipage
Solution:
abclist abc Wir zeichnen die Höhe h ein wodurch sowohl der Winkel alpha als auch die Basis b halbiert werden. Damit ist a die Hypotenuse und frac b eine Gegenkathete zum Winkel fracalpha respektive eine Ankathete zum Winkel beta. Damit folgt: sintfracalpha fracfrac ba frac &&|AN tfracalpha ^circ &&| alpha ^circ beta ^circ - fracalpha ^circ. abc Die Länge a ist a fracfrac bsintfracalpha frac.cmsin^circ .cm. abclist
Meta Information
Exercise:
minipage.textwidth Betrachte das nebenstehe gleichschenklige Dreieck mit der Basis b den Schenkeln a und den Winkeln alpha beta. abclist abc Bestimme die Winkel alpha und beta wenn bcm und acm ist. abc Bestimme die Länge a wenn alpha^circ und b.cm ist. abclist Hinweis: Zerlege das Dreieck in zwei rechtwinklige Dreiecke. minipage hfill minipage.textwidth raggedleft GleichschenkligesDreieck. minipage
Solution:
abclist abc Wir zeichnen die Höhe h ein wodurch sowohl der Winkel alpha als auch die Basis b halbiert werden. Damit ist a die Hypotenuse und frac b eine Gegenkathete zum Winkel fracalpha respektive eine Ankathete zum Winkel beta. Damit folgt: sintfracalpha fracfrac ba frac &&|AN tfracalpha ^circ &&| alpha ^circ beta ^circ - fracalpha ^circ. abc Die Länge a ist a fracfrac bsintfracalpha frac.cmsin^circ .cm. abclist
minipage.textwidth Betrachte das nebenstehe gleichschenklige Dreieck mit der Basis b den Schenkeln a und den Winkeln alpha beta. abclist abc Bestimme die Winkel alpha und beta wenn bcm und acm ist. abc Bestimme die Länge a wenn alpha^circ und b.cm ist. abclist Hinweis: Zerlege das Dreieck in zwei rechtwinklige Dreiecke. minipage hfill minipage.textwidth raggedleft GleichschenkligesDreieck. minipage
Solution:
abclist abc Wir zeichnen die Höhe h ein wodurch sowohl der Winkel alpha als auch die Basis b halbiert werden. Damit ist a die Hypotenuse und frac b eine Gegenkathete zum Winkel fracalpha respektive eine Ankathete zum Winkel beta. Damit folgt: sintfracalpha fracfrac ba frac &&|AN tfracalpha ^circ &&| alpha ^circ beta ^circ - fracalpha ^circ. abc Die Länge a ist a fracfrac bsintfracalpha frac.cmsin^circ .cm. abclist
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