randRange(3, 10) randRange(3, 10) BC AC * AC + BC * BC formattedSquareRootOf(AC * AC + BC * BC) randFromArray([ "ABC", "BAC" ]) randRange(1, 10) randRange(1, 10) (ANGLE.substring(0,1)+ANGLE.substring(2)) (function(){ if( OPPOSITE_NAME === "AC" ){ return AC; } else if( OPPOSITE_NAME === "BC" ){ return CB; } return AB_STRING; })() "AB" AB_STRING ANGLE.substring(1) (function(){ if( ADJACENT_NAME === "AC" ){ return AC; } else if( ADJACENT_NAME === "BC" ){ return BC; } return AB_STRING; })() fraction( OPPOSITE_VALUE, ADJACENT_VALUE, false, false, false, false) fractionReduce( OPPOSITE_VALUE, ADJACENT_VALUE, false, false) "\\frac{" + OPPOSITE_VALUE + "}{" + formattedSquareRootOf(AB) + "}" fractionSQRoot( OPPOSITE_VALUE , AB ) "\\frac{" + ADJACENT_VALUE + "}{" + formattedSquareRootOf(AB) + "}" fractionSQRoot( ADJACENT_VALUE , AB )

\overline{AB} = AB_STRING

\overline{OPPOSITE_NAME} = {?}

var options = {}; options[ OPPOSITE_NAME ] = "?"; options[ "AB" ] = AB_STRING; bestTriangle( BC, AC, "A", "B", "C", "", "", "", ANGLE, options ); path([ [ 0.4, 0 ], [ 0.4, 0.4 ], [ 0, 0.4 ] ]);

\displaystyle \sin( \angle ANGLE ) = SIN , \cos( \angle ANGLE ) = COS , \tan( \angle ANGLE ) = TAN

OPPOSITE_VALUE
• ADJACENT_VALUE
• AB_STRING
• WRONG_A
• WRONG_B

\overline{AB} is the hypotenuse

\overline{OPPOSITE_NAME} is opposite to \angle ANGLE

SOH CAH TOA

We know the hypotenuse and need to solve for the opposite side so we can use the sine function (SOH)

\displaystyle \sin( \angle ANGLE ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\overline{OPPOSITE_NAME}}{\overline{AB}}= \frac{\overline{OPPOSITE_NAME}}{AB_STRING}

\displaystyle \overline{OPPOSITE_NAME}=AB_STRING \cdot \sin( \angle ANGLE ) = AB_STRING \cdot SIN = OPPOSITE_VALUE

\overline{OPPOSITE_NAME} = OPPOSITE_VALUE

\overline{AB} = {?}

var options = {}; options[ OPPOSITE_NAME ] = OPPOSITE_VALUE; options[ "AB" ] = "?"; bestTriangle( BC, AC, "A", "B", "C", "", "", "", ANGLE, options ); path([ [ 0.4, 0 ], [ 0.4, 0.4 ], [ 0, 0.4 ] ]);

\displaystyle \sin( \angle ANGLE ) = SIN , \cos( \angle ANGLE ) = COS , \tan( \angle ANGLE ) = TAN

AB_STRING
• OPPOSITE_VALUE
• ADJACENT_VALUE
• WRONG_A
• WRONG_B

\overline{OPPOSITE_NAME} is the opposite to \angle ANGLE

\overline{AB} is the hypotenuse (note that it is opposite the right angle)

SOH CAH TOA

We know the opposite side and need to solve for the hypotenuse so we can use the sin function (SOH)

\displaystyle \sin( \angle ANGLE ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\overline{OPPOSITE_NAME}}{\overline{AB}} = \frac{OPPOSITE_VALUE}{\overline{AB}}

\displaystyle \overline{AB}=\frac{OPPOSITE_VALUE}{\sin( \angle ANGLE )} = \frac{OPPOSITE_VALUE}{SIN} = AB_STRING

\overline{AB} = AB_STRING

\overline{ADJACENT_NAME} = {?}

var options = {}; options[ ADJACENT_NAME ] = "?"; options[ "AB" ] = AB_STRING; bestTriangle( BC, AC, "A", "B", "C", "", "", "", ANGLE, options ); path([ [ 0.4, 0 ], [ 0.4, 0.4 ], [ 0, 0.4 ] ]);

\displaystyle \sin( \angle ANGLE ) = SIN , \cos( \angle ANGLE ) = COS , \tan( \angle ANGLE ) = TAN

ADJACENT_VALUE
• OPPOSITE_VALUE
• AB_STRING
• WRONG_A
• WRONG_B

\overline{AB} is the hypotenuse

\overline{ADJACENT_NAME} is adjacent to \angle ANGLE

SOH CAH TOA

We know the hypotenuse and need to solve for the adjacent side so we can use the cos function (CAH)

\displaystyle \cos( \angle ANGLE ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\overline{ADJACENT_NAME}}{\overline{AB}}= \frac{\overline{ADJACENT_NAME}}{AB_STRING}

\displaystyle \overline{ADJACENT_NAME}=AB_STRING \cdot \cos( \angle ANGLE ) = AB_STRING \cdot COS = ADJACENT_VALUE

\overline{ADJACENT_NAME}=ADJACENT_VALUE

\overline{AB} = {?}

var options = {}; options[ ADJACENT_NAME ] = ADJACENT_VALUE; options[ "AB" ] = "?"; bestTriangle( BC, AC, "A", "B", "C", "", "","", ANGLE, options ); path([ [ 0.4, 0 ], [ 0.4, 0.4 ], [ 0, 0.4 ] ]);

\displaystyle \sin( \angle ANGLE ) = SIN , \cos( \angle ANGLE ) = COS , \tan( \angle ANGLE ) = TAN

AB_STRING
• OPPOSITE_VALUE
• ADJACENT_VALUE
• WRONG_A
• WRONG_B

\overline{ADJACENT_NAME} is adjacent to \angle ANGLE

\overline{AB} is the hypotenuse (note that it is opposite the right angle)

SOH CAH TOA

We know the adjacent side and need to solve for the hypotenuse so we can use the cos function (CAH)

\displaystyle \cos( \angle ANGLE ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\overline{ADJACENT_NAME}}{\overline{AB}} = \frac{ADJACENT_VALUE}{\overline{AB}}

\displaystyle \overline{AB}=\frac{ADJACENT_VALUE}{\cos( \angle ANGLE )} = \frac{ADJACENT_VALUE}{COS} = AB_STRING

\overline{OPPOSITE_NAME} = OPPOSITE_VALUE

\overline{ADJACENT_NAME} = {?}

var options = {}; options[ ADJACENT_NAME ] = "?"; options[ OPPOSITE_NAME ] = OPPOSITE_VALUE; bestTriangle( BC, AC, "A", "B", "C", "", "", "" , ANGLE, options ); path([ [ 0.4, 0 ], [ 0.4, 0.4 ], [ 0, 0.4 ] ]);

\displaystyle \sin( \angle ANGLE ) = SIN , \cos( \angle ANGLE ) = COS , \tan( \angle ANGLE ) = TAN

ADJACENT_VALUE
• AB_STRING
• WRONG_A
• WRONG_B

\overline{OPPOSITE_NAME} is the opposite to \angle ANGLE

\overline{ADJACENT_NAME} is adjacent to \angle ANGLE

SOH CAH TOA

We know the opposite side and need to solve for the adjacent side so we can use the tan function (TOA)

\displaystyle \tan( \angle ANGLE ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\overline{OPPOSITE_NAME}}{\overline{ADJACENT_NAME}}= \frac{OPPOSITE_VALUE}{\overline{ADJACENT_NAME}}

\displaystyle \overline{ADJACENT_NAME}=\frac{OPPOSITE_VALUE}{\tan( \angle ANGLE )} = \frac{OPPOSITE_VALUE}{TAN} = ADJACENT_VALUE

\overline{ADJACENT_NAME} = ADJACENT_VALUE

\overline{OPPOSITE_NAME} =  ?

var options = {}; options[ OPPOSITE_NAME ] = "?"; options[ ADJACENT_NAME ] = ADJACENT_VALUE; bestTriangle( BC, AC, "A", "B", "C", "", "", "", ANGLE, options ); path([ [ 0.4, 0 ], [ 0.4, 0.4 ], [ 0, 0.4 ] ]);

\displaystyle \sin( \angle ANGLE ) = SIN , \cos( \angle ANGLE ) = COS , \tan( \angle ANGLE ) = TAN

OPPOSITE_VALUE
• AB_STRING
• WRONG_A
• WRONG_B

\overline{OPPOSITE_NAME} is the opposite to \angle ANGLE

\overline{ADJACENT_NAME} is adjacent to \angle ANGLE

SOH CAH TOA

We know the adjacent side and need to solve for the opposite side so we can use the tan function (TOA)

\displaystyle \tan( \angle ANGLE ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\overline{OPPOSITE_NAME}}{\overline{ADJACENT_NAME}}= \frac{\overline{OPPOSITE_NAME}}{ADJACENT_VALUE}

\displaystyle \overline{OPPOSITE_NAME}=ADJACENT_VALUE \cdot \tan( \angle ANGLE ) = ADJACENT_VALUE \cdot TAN = OPPOSITE_VALUE