Solution and explanation needed

skipjack

Forum Staff
Dec 2006
21,321
2,386
As sin(angle CAD) = DC/AC = (DC/MC)(MC/AC) = sin(54°)cos(24°),
angle CAD = 47.65257723° approximately.
 
Jan 2020
3
0
Texas
As sin(angle CAD) = DC/AC = (DC/MC)(MC/AC) = sin(54°)cos(24°),
angle CAD = 47.65257723° approximately.
Can you please explain further how you got that answer? The angles 54 and 24 that you are using are on the median plane, while the angle they are looking for is on the plane closest to the right.
 

skipjack

Forum Staff
Dec 2006
21,321
2,386
The required angle is angle CAD, and triangle CAD is right-angled.
DC/MC = sin(54°) as triangle DMC is right-angled.
MC/AC = cos(24°) as triangle AMC is right-angled and angle ACM = 48°/2 = 24°.
No two of the above three triangles are in the same plane.

Strictly speaking, the question doesn't give enough information, as it doesn't define the horizontal plane - it just states that AB is horizontal.
 
Jan 2020
3
0
Texas
I agree with hence why I’m curious as to how it’s 48, when there seems to be some pieces missing. Seemed a little off.
I understand the sin54, but I’m curious as to how did you know to use cos24
 

skipjack

Forum Staff
Dec 2006
21,321
2,386
It's effectively given that triangles AMC and BMC are congruent, so angle AMC is a right angle, angle ACM is 24° and cos(angle ACM) = MC/AC.