Помощ по математика( стереометрия 10 клас)

[Гост]

Здравейте, прикачих три задачи, за които имам нужда пт помощ

[ammornil]

[tex]\boxed{2}\\ V=\pi\cdot{}r^{2}\cdot{}h \Rightarrow r=\sqrt{\frac{V}{\pi\cdot{}h}} \\ \quad \\ \tg{\varphi}=\frac{h}{2\cdot{r}} \Rightarrow \varphi=\arctg{\frac{h}{2\cdot{r}}}[/tex]

Скрит текст: покажи

[tex]\varphi=\arctg{\sqrt{\frac{\pi\cdot{}h^{3}}{2\cdot{}V}}}[/tex]

[tex]\boxed{3a}\\ \angle{(BD_{1};A_{1}B_{1})}=\angle{(BD_{1};D_{1}C_{1})}=\angle{BD_{1}C_{1}} \\ \triangle{BCC_{1}}: \quad BC_{1}=\sqrt{BC^{2}+CC_{1}^{2}}=5 \\ \triangle{CA_{1}C}: \quad \angle{D_{1}C_{1}B}=90^{\circ}, \quad \tg{\angle{BD_{1}C_{1}}}=\frac{BC_{1}}{D_{1}C_{1}}=\frac{5}{3} \Rightarrow \angle{BD_{1}C_{1}}=\arctg{\frac{5}{3}}[/tex]

[tex]\boxed{3б}\\ \angle{(BD_{1};p(BCC_{1}))}=\angle{D_{1}BC_{1}} \\ \triangle{BCC_{1}}: \quad BC_{1}=\sqrt{BC^{2}+CC_{1}^{2}}=5 \\ \triangle{CA_{1}C}: \quad \angle{D_{1}C_{1}B}=90^{\circ}, \quad \tg{\angle{D_{1}BC_{1}}}=\frac{D_{1}C_{1}}{BC_{1}}=\frac{3}{5} \Rightarrow \angle{BD_{1}C_{1}}=\arctg{\frac{3}{5}}[/tex]

[tex]\boxed{3в}\\ \begin{cases} AB\bot{AD} \\ AB\bot{AA_{1}} \end{cases} \Rightarrow AB\bot{p(ADD_{1}A_{1})}[/tex]

[tex]\boxed{4}\\ ABCDM,\quad AB=BC=CD=AD=a,\quad AM=BM=CM=DM=l,\quad AC\cap{BD}=O,\quad AO=BO=CO=DO=\frac{\sqrt{2}\cdot{a}}{2},\quad MO\bot{p(ABCD)},\quad MO=H,\quad P\in{AB},\quad AP=BP \Rightarrow MP\bot{AB},\quad MP=k,\quad OP=\frac{a}{2} \\ AB=BC=CD=AD=a=6,\quad OM=H=2,\quad OA=OB=OC=OD=3\sqrt{2},\quad OP=3 \\ \begin{array}{ccl} \text{(а)} && l=\sqrt{OM^{2}+OA^{2}}=\cdots \\ \text{(б)} && k=\sqrt{OM^{2}+OP^{2}}=\cdots \\ \text{(в)} && \sin{\angle{(p(AMB), p(ABCD))}}=\frac{\normalsize{H}}{\normalsize{k}}=\cdots \\ \text{(г)} && S_{1}=a^{2}+2\cdot{a}\cdot{k}=\cdots \\ \text{(д)} && V=\frac{\normalsize{1}}{\normalsize{3}}\cdot{a^{2}}\cdot{H}=\cdots \end{array}[/tex]

[Гост]

Някой да може да помогне?

[Гост]

А, извинете не ми беше излязал отговора. Много благодаря!