Question

triunghiul dreptunghic ABC (m(∡A)=90°) este inscris in cercul de centru O si raza egala cu 24 de cm, astfel incat arcele AB si AC au masurile invers proportionale cu 3,(3) si 1,(6). Pe planul triunghiului se ridica perpendiculara AM, cu AM=36 cm. Aflati:
a) masurile arcelor AB si AC;
b) aria ΔMBC, precum si distanta de la planul A la planul (MBC)
c) masura unghiului diedru format de planele (MBC) si (ABC)

Answer 1

[tex]a)\text{Triunghiul ABC este dreptunghic si O este centrul cercului}\\ \text{circumsris,asadar O este mijlocul lui BC.}\\ \\ \{\widehat{AB},\widehat{AC}\}i.p.\{3,(3);1,(6)\}\Rightarrow \cfrac{\widehat{AB}}{\cfrac{1}{3,(3)}}=\cfrac{\widehat{AC}}{\cfrac{1}{1,(6)}}}\Rightarrow \dfrac{10\cdot \widehat{AB}}{3}=\dfrac{5\cdot \widehat{AC}}{3}=k\\ \Rightarrow \widehat{AB}=\frac{3k}{10};\widehat{AC}=\frac{3k}{5}\\ \text{Stim ca:}\\ \widehat{AB}+\widehat{AC}+\widehat{BC}=360^{\circ}\\ [/tex]
[tex]\frac{3k}{10}+\frac{3k}{5}+180^{\circ}=360^{\circ}\\ \frac{3k}{10}+\frac{3k}{5}=180^{\circ}| \cdot 10\\ 3k+6k=1800\\ 9k=1800\Rightarrow \boxed{k=200}\\ \widehat{AB}=\frac{3k}{10}\Rightarrow \boxed{\widehat{AB}=60^{\circ}}\\ \widehat{AC}=\frac{3k}{5}\Rightarrow \boxed{\widehat{AC}=120^{\circ}}\\ [/tex]
[tex]b)R=\dfrac{BC}{2}=24\Rightarrow BC=48\ cm\\ \Delta ABC\ dr.\ si\ [AO]\ median\breve{a}\Rightarrow AO=\frac{BC}{2}=24\ cm\\ Stim\ ca:m(\ \textless \ AOB)=m(\widehat{AB})=60^{\circ}\\ \hat{I}n\ \Delta AOB|m(\ \textless \ AOB)=60^{\circ}\} \\ ~~~~~~~~~~~~~~~| AO=24\ cm~~~~~~~\}\Rightarrow T.cos\Rightarrow \\ ~~~~~~~~~~~~~~~|BO=24\ cm~~~~~~~\}\\ \Rightarrow AB^2=AO^{2}+OB^{2}-2\cdot AO\cdot BO\cdot \cos\ O \\ AB^2=576+576-2\cdot 576\cdot \frac{1}{2}\\ AB^2=1152-576\\ AB^2=576\Rightarrow AB=24\ cm\\ [/tex]
[tex]\Delta ABC\ dr.|m(\ \textless \ BAC)=90^{\circ}\} \\ ~~~~~~~~~~~~~~~ |AB=24\ cm~~~~~~~\}\Rightarrow T.P.\Rightarrow AC^2=BC^2-AB^2 \\ ~~~~~~~~~~~~~~~ |BC=48\ cm~~~~~~~\}~~~~~~~~~~~~~~~ AC^2=2304-576\\ AC^2=1728\Rightarrow AC=24\sqrt{3}\ cm\\ Fie\ D\in (BC),AD\perp BC.\\ \hat{I}n\ \Delta ABC:[AD]\ \hat{l}n\breve{a}ltime\Rightarrow AD=\dfrac{AB\cdot AC}{BC}\\ AD=\dfrac{24\cdot 24\sqrt3}{48}\Rightarrow AD=12\sqrt3\ cm\\ [/tex]
[tex]MA\perp (ABC)\}\\ ~~~~~~~~~~~~~~~~~~~~\}\Rightarrow MA\perp AD~~~~~~~~~\}\\ AD\subset(ABC)~\}~~~~~~~~~~~~~~~~~~~~~~~~~~~~\}\Rightarrow MD\perp BC\ (T3\perp)\\ ~~~~~~~~~~~~~~~~~~~~D\in (BC),AD\perp BC~ \}\\ \Delta MAD\ dr.|m(\ \textless \ A)=90^{\circ}~~\}\\ ~~~~~~~~~~~~~~~~|AD=12\sqrt 3\ cm\}\Rightarrow T.P.\Rightarrow MD^2=MA^2+AD^2\\ ~~~~~~~~~~~~~~~~|MA=36\ cm~~~\}~~~~~~~~~~~~~~~MD^2=1296+432\\ MD^2=1728\Rightarrow MD=24\sqrt{3}\ cm\\ A_{\Delta{MBC}}=\dfrac{MD\cdot BC}{2} [/tex]
[tex]A_{\Delta{MBC}}=\frac{24\sqrt3\cdot 48}{2}=576\sqrt 3\ cm^2\\ \text{Distanta de la A la planul MBC o aflii cu teorema inaltimii in }\\ \text{triunghiul MAD.}\\ \\ c)MD\perp BC,MD\subset(MBC)\}\\ ~~~~AD\perp BC,AD\subset(ABC)~~\}\Rightarrow m(\ \textless \ ((ABC);(MBC)))=\\ ~~~~ (ABC)\cap (MBC)=BC~~~\}~~~~~~~=m(\ \textless \ ADM)\\ In\ \Delta {MAD}\ dr.| MD=24\sqrt3\ cm\}\\ ~~~~~~~~~~~~~~~~~~~~~|AD=12\sqrt 3\ cm~\}\Rightarrow \cos D=\frac{AD}{MD}\\ ~~~~~~~~~~~~~~~~~~~~~| m(\ \textless \ A)=90^{\circ}~~~\}~~~~\cos D=\frac{12\sqrt{3}}{24\sqrt3}[/tex]
$\cos D=\frac{1}{2}\Rightarrow \boxed{m(\ \textless \ ADM)=60^{\circ}}$