Category: Geometry

Question Number 195287 by Mingma last updated on 29/Jul/23 Answered by MM42 last updated on 29/Jul/23 $$CD=DE\Rightarrow \mathrm{\angle }C2=\mathrm{\angle }E\mathrm{\&}\mathrm{\angle }D1+\mathrm{\angle }E=60$$$$\mathrm{\angle }C1=60-\mathrm{\angle }C2=60-\mathrm{\angle }E=\mathrm{\angle }D1$$$$\frac{CD}{Sin60}=\frac{2}{SinC1}\mathrm{\&}\frac{DE}{Sin120}=\frac{BE}{SinD1}$$$$\frac{2}{SinC1}=\frac{BE}{SinD1}\Rightarrow BE=2✓$$…

Question Number 194998 by a.lgnaoui last updated on 22/Jul/23 $$\mathrm{Resolution}\mathrm{du}\mathrm{probldme}\mathrm{pose}\mathrm{par}$$$$\mathrm{sonukgindia}(16.7.2023)$$$$\mathrm{voir}\mathrm{Q}194819$$$$△\mathbf{ABC}\mathbf{AM}=\mathbf{AN}=\mathbf{AD}\cos \frac{\mathit{\alpha }}{2}$$$$\{\begin{array}{l}\mathbf{AC}=\mathbf{AM}+\mathbf{MC}=17\left(1\right)\\ \mathbf{AB}=\mathbf{AN}+\mathbf{NB}=18\left(2\right)\end{array}$$$$\mathbf{AB}-\mathbf{AC}=1=\mathbf{NB}-\mathbf{MC}\left(3\right)$$$$△\mathbf{CDE}\cos \frac{\mathbf{C}}{2}=\frac{\mathbf{CM}}{\mathbf{CD}}=\frac{\mathbf{CE}}{\mathbf{CD}}\Rightarrow \mathbf{CM}=\mathbf{CE}$$$$\bigtriangleup\boldsymbol{\mathrm{BDE}}\:\:\:\mathrm{cos}\:\frac{\boldsymbol{\mathrm{B}}}{\mathrm{2}}=\frac{\boldsymbol{\mathrm{BE}}}{\boldsymbol{\mathrm{BD}}}=\frac{\boldsymbol{\mathrm{BN}}}{\boldsymbol{\mathrm{BD}}}\Rightarrow\boldsymbol{\mathrm{BN}}=\boldsymbol{\mathrm{BE}} \…

Question Number 195046 by Mathstar last updated on 22/Jul/23 Commented by Mathstar last updated on 23/Jul/23 $$$$$$\mathrm{Given}\mathrm{curve}\mathrm{ysin}\left(\mathrm{x}\right)\mathrm{and}\mathrm{a}\mathrm{green}$$$$\mathrm{square};\mathrm{solve}\mathrm{for}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{square}.$$ Answered by…

Question Number 194938 by Mingma last updated on 20/Jul/23 Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 194847 by cortano12 last updated on 17/Jul/23 $$\underbrace{}$$ Commented by Tinku Tara last updated on 17/Jul/23 $$z=?,\mathrm{without}\mathrm{knowing}z\mathrm{you}\mathrm{cannot}$$$$\mathrm{find}n$$…