- #1
MHB Area of Triangle ABC Given Dimensions
- Thread starter maxkor
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In triangle ABC, given the dimensions AC=BD, CE=2, ED=1, AE=4, and the angle relationship where angle CAE is twice angle DAB, the area of triangle ABC is being calculated. The discussion emphasizes the use of the law of cosines to find angles and side lengths, leading to the determination of angle alpha and the lengths AD and AB. There is a mention of using Geogebra for visual representation, but concerns are raised about the accuracy of the diagram. The calculations yield specific angle measures and side lengths, which are crucial for finding the area. The conversation highlights the importance of accurate diagrams and calculations in solving geometric problems.
- #2
jonah1
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Beer induced request follows.
We can't help you if we don't where you are stuck.
Please show us what you have tried and exactly where you are stuck.maxkor said:
We can't help you if we don't where you are stuck.
- #3
- #4
HOI
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How did you get those?
- #5
maxkor
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Geogebra...
- #6
mrtwhs
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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 $\alpha \approx 71.170769^{\circ}$.
Your diagram is incorrect.
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 $\alpha \approx 71.170769^{\circ}$.
Your diagram is incorrect.
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