Question Number 206971 by MATHEMATICSAM last updated on 02/May/24
$$\mathrm{Construct}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{whose}\:\mathrm{sine}\:\mathrm{is} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}\:+\:\sqrt{\mathrm{5}}}\:. \\ $$
Answered by Rasheed.Sindhi last updated on 02/May/24
$$\mathrm{sin}\theta=\:\frac{\mathrm{3}}{\mathrm{2}\:+\:\sqrt{\mathrm{5}}} \\ $$$$\bullet\theta\:{is}\:{an}\:{angle}\:{of}\:{a}\:{right}\:{triangle} \\ $$$${whose}\:{hypotenuse}\:{is}\:\mathrm{2}+\sqrt{\mathrm{5}}\:{and} \\ $$$${opposite}\:{side}\:{is}\:\mathrm{3} \\ $$$$\bullet\:\sqrt{\mathrm{5}}\:{is}\:{a}\:{hypotenuse}\:{of}\:{a}\:{right}\: \\ $$$${triangle}\:{having}\:{legs}\:\mathrm{2}\:{and}\:\mathrm{1} \\ $$$$\blacktriangleright\mathcal{D}{raw}\:{a}\:{right}\:{triangle}\:{having} \\ $$$${legs}\:\mathrm{2}\:{and}\:\mathrm{1}.\:{Extend}\:{its}\:{hypotenuse} \\ $$$$\sqrt{\mathrm{5}}\:\:{to}\:\sqrt{\mathrm{5}}\:+\mathrm{2}. \\ $$$$\mathcal{D}{raw}\:{semicircle}\:{on}\:{line}\:{segment} \\ $$$$\overline {\boldsymbol{\mathrm{AB}}}\:{having}\:{measre}\:\sqrt{\mathrm{5}}\:+\mathrm{2} \\ $$$$\mathcal{C}{ut}\:{off}\:\mathrm{3}\:{units}\:{from}\:\boldsymbol{\mathrm{B}}\:{of}\:{semicircle} \\ $$$${at}\:\boldsymbol{\mathrm{C}}. \\ $$$${Join}\:\boldsymbol{{C}}\:{to}\:\boldsymbol{\mathrm{A}}\:{and}\:\boldsymbol{\mathrm{B}} \\ $$$$\angle\boldsymbol{\mathrm{BAC}}\:{is}\:{a}\:{required}\:{angle}. \\ $$
Commented by Rasheed.Sindhi last updated on 02/May/24
Commented by MATHEMATICSAM last updated on 02/May/24
$$\mathrm{Thank}\:\mathrm{you}! \\ $$