The questions I am stumped on are parts c) and d). I am bewildered as to which triangles I am meant to use. For context, I have got an answer of 54.7 for part b and this is correct.
Solutions to c and d are below:
Solutions
c) 35 degrees d) 24 degrees
I would really appreciate any help about where I am going wrong here. Thank you.
In cross section through points A, F and M by a plane containing these points, is triangle AMF which gives first angle $\alpha$:
$\tan (\alpha)=\frac{15}{10}=1.4\Rightarrow \alpha=56.3$
In cross section through points A, B and M by a plane containing these points, is triangle AMB which gives first angle $\beta$:
$\tan (\beta)=\frac{15}{\sqrt{10^2+10^2}}=1.06 \Rightarrow \beta= 46.7$
Second question:
a):
$AG=\sqrt{AB^2+BC^2+CG^2}=\sqrt{4^2+4^2+8^2}=9.7$
b):Consider a plane contanig vertices A,C,E and G. we have right angled triangles AGC and AGE.
$tan(\gamma)=\frac8{\sqrt{4^2+4^2}}=1.41\Rightarrow \gamma=54.7$
c):Angle between AG and EFGH is equale to angle between AG and ABCD:
$\theta=\gamma=54.7$
d):
$AH^2=AE^2+EH^2=4^2+8^2=80 \Rightarrow AH=4\sqrt 5$
$\tan (\alpha)=\frac 4{4\sqrt 5}=0.447\Rightarrow \alpha= 24^o$