shuffle(randFromArray([[6,8], [8, 15], [10, 24], [12,16], [14, 48], [20, 48], [16, 30]])) BC BC sqrt(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; })() "AB" AB ANGLE.substring(1) (function(){ if( ADJACENT_NAME === "AC" ){ return AC; } else if( ADJACENT_NAME === "BC" ){ return BC; } return AB; })() fraction( reduce(OPPOSITE_VALUE, ADJACENT_VALUE)[0], reduce(OPPOSITE_VALUE, ADJACENT_VALUE)[1] , false, false, false, false) OPPOSITE_VALUE / ADJACENT_VALUE fraction(reduce(OPPOSITE_VALUE, AB)[0], reduce(OPPOSITE_VALUE, AB)[1]) OPPOSITE_VALUE / AB fraction(reduce(ADJACENT_VALUE, AB)[0], reduce(ADJACENT_VALUE, AB)[1]) ADJACENT_VALUE / AB

\overline{AB} = AB

\overline{OPPOSITE_NAME} = {?}

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

\displaystyle \sin( \angle ANGLE ) = SIMPLE_SIN , \cos( \angle ANGLE ) = SIMPLE_COS , \tan( \angle ANGLE ) = SIMPLE_TAN

OPPOSITE_VALUE

\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}

Since we have already been given \sin( \angle ANGLE ), we can set up a proportion to find \overline{OPPOSITE_NAME}.

\displaystyle \sin( \angle ANGLE ) = SIMPLE_SIN = \frac{\overline{OPPOSITE_NAME}}{AB}

Simplify.

\overline{OPPOSITE_NAME} = 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 ) = SIMPLE_SIN , \cos( \angle ANGLE ) = SIMPLE_COS , \tan( \angle ANGLE ) = SIMPLE_TAN

AB

\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}}

Since we have already been given \sin( \angle ANGLE ), we can set up a proportion to find \overline{AB}.

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

Simplify.

\overline{AB} = AB

\overline{AB} = AB

\overline{ADJACENT_NAME} = {?}

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

\displaystyle \sin( \angle ANGLE ) = SIMPLE_SIN , \cos( \angle ANGLE ) = SIMPLE_COS , \tan( \angle ANGLE ) = SIMPLE_TAN

\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}

Since we have already been given \cos( \angle ANGLE ), we can set up a proportion to find \overline{ADJACENT_NAME}.

\displaystyle \cos( \angle ANGLE ) = SIMPLE_COS = \frac{\overline{ADJACENT_NAME}}{AB}

Simplify.

\overline{ADJACENT_NAME} = 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 ) = SIMPLE_SIN , \cos( \angle ANGLE ) = SIMPLE_COS , \tan( \angle ANGLE ) = SIMPLE_TAN

AB

\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}}

Since we have already been given \cos( \angle ANGLE ), we can set up a proportion to find \overline{AB}.

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

Simplify.

\overline{AB} = AB

\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 ) = SIMPLE_SIN , \cos( \angle ANGLE ) = SIMPLE_COS , \tan( \angle ANGLE ) = SIMPLE_TAN

\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}}

Since we have already been given \tan( \angle ANGLE ), we can set up a proportion to find \overline{ADJACENT_NAME}.

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

Simplify.

\overline{ADJACENT_NAME} = 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 ) = SIMPLE_SIN , \cos( \angle ANGLE ) = SIMPLE_COS , \tan( \angle ANGLE ) = SIMPLE_TAN

OPPOSITE_VALUE

\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}

Since we have already been given \tan( \angle ANGLE ), we can set up a proportion to find \overline{OPPOSITE_NAME}.

\displaystyle \tan( \angle ANGLE ) = SIMPLE_TAN = \frac{\overline{OPPOSITE_NAME}}{ADJACENT_VALUE}

Simplify.

\overline{OPPOSITE_NAME} = OPPOSITE_VALUE