Änderungen von Dokument Lösung Lineare Algebra

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  • DreieckSkizze.svg

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Inhalt

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1	1	=== Teilaufgabe a) ===
2	2	{{detail summary="Erwartungshorizont"}}
	3	+[[image:Lösunga).png||width="220" style="float: right"]]
3	3	{{formula}}
4	4	\begin{align*}
5	5	&A_1(3|3|0), \ A_2(-3|3|0), \\
...	...	@@ -7,7 +7,7 @@
7	7	& C'_1(3|1|0), C'_2(1|3|0)
8	8	\end{align*}
9	9	{{/formula}}
10		-[[image:Lösunga).png||width="180"]]
	11	+
11	11	{{/detail}}
12	12
13	13
...	...	@@ -27,7 +27,7 @@
27	27	\Bigl|\overrightarrow{C_1C_2}\Bigl|=\Bigl|\overrightarrow{B_1C_1}\Bigl|=\Bigl|\overrightarrow{B_1C_2}\Bigl|=2\sqrt2
28	28	{{/formula}}
29	29	<br>
30		-Damit ist das Dreieck {{formula}}C_1B_1C_2{{/formula}} gleichseitig.
	31	+Damit ist das Dreieck {{formula}}B_2C_1C_2{{/formula}} gleichseitig.
31	31	<p></p>
32	32	Mittelpunkt der Strecke {{formula}}C_1C_2{{/formula}}:
33	33	{{formula}}
...	...	@@ -102,7 +102,7 @@
102	102
103	103	{{formula}}
104	104	\cos(\varphi)=\frac{\text{Kirchturmhöhe} \ h}{\text{gesuchter Abstand}\ a} \ \Leftrightarrow \
105		-a=\frac{h}{\cos(\varphi)}\approx\frac{30}{\cos(42{,}0^\circ)}\approx40{,}4
	106	+a=\frac{h}{\cos(\varphi)}\approx\frac{30}{\cos(42^\circ)}\approx40{,}4
106	106	{{/formula}}
107	107	<p></p>
108	108	Der Abstand der Turmspitze und ihrem Schattenpunkt beträgt ca. 40,4 Meter.

DreieckSkizze.svg

Author

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1		-XWiki.akukin

Größe

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1		-8.4 KB

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