You should use the law of cosines on triangle ECD to find CD, and then use the law of cosines on triangle ACD to find AD, and then use the law of cosines on triangle ABD to find AB.In summary, the conversation discussed solving for the area of a triangle given certain side lengths and angles. The law of cosines was used to find the missing side lengths, and the final result was obtained using the law of cosines on the remaining triangle.
#1
maxkor
84
0
In triangle ABC $AC=BD, CE=2, ED=1, AE=4$ and $\angle CAE=2 \angle DAB$. Find area ABC.
In triangle ABC $AC=BD, CE=2, ED=1, AE=4$ and $\angle CAE=2 \angle DAB$. Find area ABC.
Please show us what you have tried and exactly where you are stuck.
We can't help you if we don't where you are stuck.
#3
maxkor
84
0
This is what it looks like, but how to justify the red ones or maybe there is another way?
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#4
HOI
921
2
How did you get those?
#5
maxkor
84
0
Geogebra...
#6
mrtwhs
47
0
In your diagram, all three sides of $\triangle AEC$ are known. Using the law of cosines you get $2\alpha = \cos^{-1}(7/8)$ and $\alpha \approx 14.47751219^{\circ}$.
Continuing to use the law of cosines we get $AD=\dfrac{3\sqrt{10}}{2}$ and $AB=3\sqrt{6}$.
Finally, using the law of cosines on $\triangle ABD$ we get $\alpha \approx 71.170769^{\circ}$.