Егер sinα=35 болса, α сүйір бұрышы үшін cosα, tgα− ны анықтаңыз. это БЖБ

begin{gathered}(x-7)^{2} -49=0;\\(x-7)^{2} =49;\\\left [ \begin{array}{lcl} {{x-7=-7,} \\ {x-7=7;}} \end{array} \right.\Leftrightarrow\left [ \begin{array}{lcl} {{x=-7+7,} \\ {x=7+7;}} \end{array} \right.\Leftrightarrow \left [ \begin{array}{lcl} {{x=0,} \\ {x=14.}} \end{array} \right.\end{gathered}

ответ : 0;14.

\begin{gathered}(6+y) ^{2} -81=0;\\(6+y) ^{2} -9^{2} =0;\\(6+y-9)(6+y+9)=0;\\(y-3)(y+15)=0;\\\left [ \begin{array}{lcl} {{y-3=0,} \\ {y+15=0;}} \end{array} \right.\Leftrightarrow\left [ \begin{array}{lcl} {{y=3,} \\ {y=-15.}} \end{array} \right.\end{gathered}

ответ : -15; 3.

\begin{gathered}100- (z-19)^{2} =0;\\ (z-19)^{2} = 10^{2} \\\left [ \begin{array}{lcl} {{z-19=10,} \\ {z-19=-10;}} \end{array} \right.\Leftrightarrow\left [ \begin{array}{lcl} {{z=29,} \\ {z=9.}} \end{array} \right.\end{gathered}

ответ : 9;29.

\begin{gathered}25- (13+y)^{2} =0;\\(13+y)^{2} = 5^{2} ;\\\left [ \begin{array}{lcl} {{13+y=-5,} \\ {13+y=5;}} \end{array} \right.\Leftrightarrow\left [ \begin{array}{lcl} {{y=-18,} \\ {y=-8.}} \end{array} \right.\end{gathered}