In a triangle with corners A, B and C, all sides are 9 cm. point D is located such that the line AD is normal (ie perpendicular) to the plane of the triangle (ie surface) and the distance AD is 12 cm. Calculate the acute angle between lines AC and BD.
Using the information in above also calculate the acute angle between the ABC and BCD planes.
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Ok so. For the first part, I believe I got 56.99 degrees. I'm not sure how correct this is (Got 15cm for the height from the middle of the plane to the top, then hieght base from apex is 14.31 for the larger triangle and 7.79 for the smaller.) Then the angle is 56.99)
I'm not sure how to use this information to get the rest, aka 'the acute angle between the ABC and BCD planes.' (Heck I'm not sure if this is even all correct, I'm struggling with isometrics atm)
Since AD⊥ABC, AD⊥AC
So lines AC and BD do not intersect.
∆ABD is a right triangle, with tanB = 12/9
What's amiss here?
BD = 15, but that's not an altitude of anything.
Let E be the midpoint of BC.
The altitude of ∆ABC EA=9√3/2, so the dihedral angle x=angle AED between the two triangles is tanx = 12/(9√3/2). That is the 56.99° angle.