Question Number 203321 by SonGoku last updated on 16/Jan/24 $$\mathrm{Help}-\mathrm{me}! \\ $$$$\: \\ $$$$\mathrm{Observe}\:\mathrm{points}\:\mathrm{A},\:\mathrm{B}\:\mathrm{and}\:\mathrm{C}\:\mathrm{below}\:\mathrm{and}\:\mathrm{find}\:\mathrm{the}\:\mathrm{widthof}\:\mathrm{a}\:\mathrm{lake}\:\mathrm{according}\:\mathrm{to}\:\mathrm{the}\:\mathrm{following}\:\mathrm{data}: \\ $$$$\: \\ $$$$\left(\mathrm{AB}\right)\mathrm{m};\:\hat {\mathrm{C}}\:=\:\mathrm{39}°\mathrm{52}'\mathrm{12}'' \\ $$$$\left(\mathrm{BC}\:−\:\mathrm{257}.\mathrm{5}\right)\mathrm{m};\:\hat {\mathrm{A}}\:=\:\mathrm{97}°\mathrm{7}'\mathrm{56}'' \\ $$$$\left(\mathrm{CA}\:−\:\mathrm{30}\right)\mathrm{m};\:\hat {\mathrm{B}}\:=\:\mathrm{42}°\mathrm{59}'\mathrm{52}''…
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Question Number 203244 by cortano12 last updated on 13/Jan/24 Answered by MathematicalUser2357 last updated on 14/Jan/24 $$\mathrm{If}\:\mathrm{you}\:\mathrm{took}\:\mathrm{that}\:\mathrm{image}\:\mathrm{on}\:\mathrm{FB}, \\ $$$$\mathrm{You}\:\mathrm{will}\:\mathrm{be}\:\mathrm{known}\:\mathrm{soon}. \\ $$$$\mathrm{Here}\:\mathrm{is}\:\mathrm{a}\:\mathrm{proof}: \\ $$ Commented by…
Question Number 203171 by a.lgnaoui last updated on 11/Jan/24 $$\mathrm{Ratio}\:\mathrm{of}\:\mathrm{green}\:\mathrm{blue}\:\mathrm{and}\:\mathrm{Red}\:\mathrm{triangles}? \\ $$$$\mathrm{Evaluate}\:\mathrm{these}\:\mathrm{Area}\:\mathrm{for}\:\boldsymbol{\alpha}=\mathrm{25}°\:\:\mathrm{and} \\ $$$$\mathrm{length}\:\mathrm{of}\:\mathrm{aquart}=\mathrm{1} \\ $$ Commented by a.lgnaoui last updated on 11/Jan/24 Commented by…
Question Number 202976 by ajfour last updated on 06/Jan/24 Commented by ajfour last updated on 06/Jan/24 $${If}\:\mathrm{tan}\:\alpha={t}\:,\:{find}\:{h}. \\ $$ Answered by mr W last updated…
Question Number 202614 by a.lgnaoui last updated on 30/Dec/23 $$\mathrm{find}\:\mathrm{x} \\ $$$$ \\ $$ Commented by a.lgnaoui last updated on 30/Dec/23 Commented by a.lgnaoui last…
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Question Number 201854 by cortano12 last updated on 14/Dec/23 Commented by cortano12 last updated on 14/Dec/23 $$\:\:\begin{cases}{\mathrm{x}^{\mathrm{2}} −\mathrm{3y}^{\mathrm{2}} =\frac{\mathrm{17}}{\mathrm{x}}}\\{\mathrm{3x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} =\frac{\mathrm{23}}{\mathrm{y}}}\end{cases} \\ $$$$\:\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\:\sqrt[{\mathrm{m}}]{\mathrm{n}}\:,\:\mathrm{m},\mathrm{n}\:\in\mathbb{Z}^{+}…
Question Number 201425 by mathlove last updated on 06/Dec/23 $${prove}\:{that} \\ $$$$\frac{\mathrm{1}−{cosA}+{cosB}−{cos}\left({A}+{B}\right)}{\mathrm{1}+{cosA}−{cosB}−{cos}\left({A}+{B}\right)}={tan}\frac{{A}}{\mathrm{2}}\centerdot{cot}\frac{{B}}{\mathrm{2}} \\ $$ Answered by esmaeil last updated on 06/Dec/23 $$\frac{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{A}+{B}}{\mathrm{2}}\right)\:−\mathrm{2}{sin}\left(\frac{{B}+{A}}{\mathrm{2}}\right){sin}\left(\frac{{B}−{A}}{\mathrm{2}}\right)}{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{B}+{A}}{\mathrm{2}}\right)−\mathrm{2}{sin}\left(\frac{{B}+{A}}{\mathrm{2}}\right){sin}\left(\frac{{A}−{B}}{\mathrm{2}}\right)}= \\…
Question Number 201426 by mathlove last updated on 06/Dec/23 $${cosx}−\sqrt{\mathrm{3}}{sinx}=\mathrm{1} \\ $$$${x}=? \\ $$ Answered by cortano12 last updated on 06/Dec/23 $$\:\:\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{cos}\:\mathrm{x}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{sin}\:\mathrm{x}\right)=\:\mathrm{1} \\ $$$$\:\:\:\:\mathrm{cos}\:\left(\mathrm{x}+\mathrm{60}°\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}=\:\mathrm{cos}\:\mathrm{60}°\: \\…