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GeometryQuestion and Answers: Page 78

Question Number 75222    Answers: 0   Comments: 1

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Question Number 75058    Answers: 1   Comments: 2

in triangle: ABC: a=(√(2 )),b−c=(((√2)+1)/2),B^ −C^ =(𝛑/2) find: h_a , S_(ABC ) ,d_(a ) , R ,A^ .

$$\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{triangle}}:\:\:\boldsymbol{\mathrm{ABC}}: \\ $$$$\boldsymbol{\mathrm{a}}=\sqrt{\mathrm{2}\:},\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{c}}=\frac{\sqrt{\mathrm{2}}+\mathrm{1}}{\mathrm{2}},\overset{} {\boldsymbol{\mathrm{B}}}−\overset{} {\boldsymbol{\mathrm{C}}}=\frac{\boldsymbol{\pi}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{find}}:\:\:\boldsymbol{\mathrm{h}}_{\boldsymbol{\mathrm{a}}} ,\:\:\boldsymbol{\mathrm{S}}_{\boldsymbol{\mathrm{ABC}}\:\:} ,\boldsymbol{\mathrm{d}}_{\boldsymbol{\mathrm{a}}\:\:\:} ,\:\boldsymbol{\mathrm{R}}\:\:\:\:,\overset{} {\boldsymbol{\mathrm{A}}}. \\ $$

Question Number 74748    Answers: 1   Comments: 3

Question Number 74747    Answers: 1   Comments: 0

Question Number 74713    Answers: 1   Comments: 1

Question Number 74726    Answers: 1   Comments: 3

Question Number 74778    Answers: 0   Comments: 0

Question Number 74697    Answers: 1   Comments: 0

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Question Number 74509    Answers: 0   Comments: 4

Question Number 74240    Answers: 1   Comments: 3

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Question Number 74112    Answers: 1   Comments: 0

Question Number 74087    Answers: 0   Comments: 15

(Q73828) prove that no cube exists whose corners are located on all faces of an other cube.

$$\left({Q}\mathrm{73828}\right) \\ $$$${prove}\:{that}\:{no}\:{cube}\:{exists}\:{whose}\:{corners} \\ $$$${are}\:{located}\:{on}\:{all}\:{faces}\:{of}\:{an}\:{other}\:{cube}. \\ $$

Question Number 73884    Answers: 0   Comments: 0

Question Number 73828    Answers: 1   Comments: 3

Question Number 73816    Answers: 0   Comments: 8

Question Number 73800    Answers: 0   Comments: 8

Please draw the shape and find the angles QR = 6 cm RS = 7 cm PS = 4 cm

$$\mathrm{Please}\:\mathrm{draw}\:\mathrm{the}\:\mathrm{shape}\:\mathrm{and}\:\mathrm{find}\:\mathrm{the}\:\mathrm{angles} \\ $$$$\mathrm{QR}\:\:\:=\:\:\mathrm{6}\:\:\mathrm{cm} \\ $$$$\mathrm{RS}\:\:\:=\:\:\mathrm{7}\:\mathrm{cm} \\ $$$$\mathrm{PS}\:\:=\:\:\mathrm{4}\:\mathrm{cm} \\ $$

Question Number 73737    Answers: 2   Comments: 1

Question Number 73673    Answers: 2   Comments: 3

Question Number 73663    Answers: 1   Comments: 0

Question Number 73620    Answers: 1   Comments: 0

Question Number 73503    Answers: 1   Comments: 2

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