Given the following angles:
\overline{AB} \perp \overline{CD},
line segments AB and CD are perpendicular.
\green{\angle{CGE}} = ACCUTEANGLE^\circ
\green{\angle{AGF}} = 90 - ACCUTEANGLE^\circ
\green{\angle{DGF}} = ACCUTEANGLE^\circ
What is
\blue{\angle{AGF}} = {?}
\blue{\angle{CGE}} = {?}
\blue{\angle{BGE}} = {?}
NOTE: Angles not necessarily drawn to scale.
\Large{{}^\circ}
Because we know \overline{AB} \perp \overline{CD}, we know
\purple{\angle{CGB}} = 90^\circ
label( [2.2, 1.7], "\\purple{90^\\\circ}",
"above right" );
arc( [0, 0], 3, 0, 90, { stroke: PURPLE } );
\orange{\angle{EGB}} = \green{\angle{AGF}}
= 90 - ACCUTEANGLE^\circ,
because they are opposite angles from each other. Opposite angles
are congruent (equal).
label( [1.2, 0], "\\orange{" + (90 - ACCUTEANGLE) + "^\\circ}",
"above right" );
arc( [0, 0], 1.2, 0, 68, { stroke: ORANGE } );
Because we know \overline{AB} \perp \overline{CD}, we know
\purple{\angle{AGD}} = 90^\circ
label( [-2.2, -1.7], "\\purple{90^\\circ}",
"below left" );
arc( [0, 0], 3, 180, 270, { stroke: PURPLE } );
\orange{\angle{EGB}} = \purple{90^\circ}
- \green{\angle{CGE}} = 90 - ACCUTEANGLE^\circ
label( [1.2, 0], "\\orange{" + (90 - ACCUTEANGLE) + "^\\circ}",
"above right" );
arc( [0, 0], 1.2, 0, 68, { stroke: ORANGE } );
Because we know \overline{AB} \perp \overline{CD}, we know
\purple{\angle{CGB}} = 90^\circ
label( [2.2, 1.7], "\\purple{90^\\circ}",
"above right" );
arc( [0, 0], 3, 0, 90, { stroke: PURPLE } );
\orange{\angle{AGF}} =
\purple{90^\circ} - \green{\angle{DGF}} =
90 - ACCUTEANGLE^\circ
label( [-1.2, 0], "\\orange{" + (90 - ACCUTEANGLE) + "^\\circ}",
"below left" );
arc( [0, 0], 1.2, 180, 248, { stroke: ORANGE } );
\blue{\angle{AGF}} = \orange{\angle{EGB}} =
90 - ACCUTEANGLE^\circ,
because they are opposite from each other. Opposite angles are
congruent (equal).
ORIGINAL_LABEL.remove();
label( [-1.2, -0.75],
"\\blue{\\angle{AGF}}=" + (90 - ACCUTEANGLE) + "^\\circ",
"below left" );
\blue{\angle{CGE}} =
\purple{90^\circ} - \orange{\angle{EGB}} =
ACCUTEANGLE^\circ
ORIGINAL_LABEL.remove();
label( [0.5, 1.8],
"\\blue{\\angle{CGE}} = " + ACCUTEANGLE + "^\\circ",
"above" )
\blue{\angle{BGE}} = \orange{\angle{AGF}} =
90 - ACCUTEANGLE^\circ,
because they are opposite from each other. Opposite angles are
congruent (equal).
ORIGINAL_LABEL.remove();
label( [1.5, 0],
"\\blue{\\angle{BGE}} = " + (90 - ACCUTEANGLE) + "^\\circ",
"above right" );
Given the following:
\purple{\angle{ABC}} = Tri_Z^\circ\green{\angle{ACB}} = Tri_Y^\circWhat is \blue{\angle{DAB}}?
\purple{\angle{ABC}} = Tri_Z^\circ\green{\angle{DAB}} = 180 - Tri_X^\circWhat is \blue{\angle{ACB}}?
NOTE: Angles not necessarily drawn to scale.
\Large{{}^\circ}
\orange{\angle{BAC}} =
180^\circ - \purple{\angle{ABC}} - \green{\angle{ACB}} =
180 - Tri_Y - Tri_Z^\circ
,
This is because angles inside a triangle add up to 180^\circ.
label( [-3.3, -2], "\\orange{" + Tri_X + "^\\circ}",
"above right" );
arc( [-4, -2], 0.75, 0, 49, {stroke: ORANGE} );
\orange{\angle{BAC}} = 180^\circ - \green{\angle{DAB}} = 180 - Tri_Y - Tri_Z^\circ,
because supplementary angles along a line add up to 180^\circ.
label( [-3.3, -2], "\\orange{" + Tri_X + "^\\circ}",
"above right" );
arc( [-4, -2], 0.75, 0, 49, {stroke: ORANGE} );
\blue{\angle{DAB}} = 180^\circ - \orange{\angle{BAC}} = Tri_Y + Tri_Z^\circ,
because supplementary angles along a line add up to 180^\circ.
ORIGINAL_LABEL.remove();
label( [-4.7, -2],
"\\blue{\\angle{DAB}} = " + (Tri_Y + Tri_Z) + "^\\circ",
"above left" );
\blue{\angle{ACB}} = 180^\circ - \orange{\angle{BAC}} - \purple{\angle{ABC}} = Tri_Y^\circ,
because angles inside a triangle add up to 180^\circ.
ORIGINAL_LABEL.remove();
label( [2.80, -2],
"\\blue{\\angle{ACB}} = " + Tri_Y + "^\\circ",
"above left" );
Given the following:
\overline{HI} \parallel \overline{JK},
line segments HI and JK are parallel.\purple{\angle{BAC}} = Tri_X^\circ
\purple{\angle{AKJ}} = Tri_Y^\circ
\green{\angle{AJK}} = Tri_Z^\circ
\green{\angle{AHI}} = Tri_Z^\circ
What is
\blue{\angle{AIH}} = {?}
\blue{\angle{AKJ}} = {?}
\blue{\angle{BAC}} = {?}
NOTE: Angles not necessarily drawn to scale.
\Large{{}^\circ}
\orange{\angle{AHI}} = \green{\angle{AJK}},
because they are corresponding angles formed by two parallel lines and
a transversal line. Corresponding angles are congruent (equal).
label( [-4.60, 0.75], "\\orange{" + Tri_Z + "^\\circ}",
"below" );
arc( [-5.07, 1.75], 1, 260, 325, {stroke: ORANGE} );
\orange{\angle{AJK}} = \green{\angle{AHI}},
because they are corresponding angles formed by two parallel lines and
a transversal line. Corresponding angles are congruent (equal).
label( [-4.00, 4.25], "\\orange{" + Tri_Z + "^\\circ}",
"below" );
arc( [-4.47, 5.25], 1, 257, 325, {stroke: ORANGE} );
\blue{\angle{AIH}} =
180^\circ - \orange{\angle{AHI}} - \purple{\angle{BAC}} =
180 - Tri_X - Tri_Z^\circ
,
because the three angles are contained in \triangle{AHI}.
Angles inside a triangle add up to 180^\circ.
ORIGINAL_LABEL.remove();
label( [0, -2.50],
"\\blue{\\angle{AIH}} = " + (180 - Tri_X - Tri_Z) + "^\\circ",
"left" );
ORIGINAL_LABEL.remove();
if ( RAND3 === 1 ) {
label( [3.3, -2.6],
"\\blue{\\angle{AKJ}} = " + Tri_Y + "^\\circ",
"above" );
} else {
label( [-5.5, -3.5],
"\\blue{\\angle{BAC}} = " + Tri_X + "^\\circ",
"above right" );
}
\blue{\angle{AKJ}} =
180^\circ - \orange{\angle{AJK}} - \purple{\angle{BAC}} =
Tri_Y^\circ
\blue{\angle{BAC}} =
180^\circ - \orange{\angle{AJK}} - \purple{\angle{AKJ}} =
Tri_X^\circ
,
because the three angles are contained in \triangle{AJK}.
Angles inside a triangle add up to 180^\circ.
Given the following:
\overline{DE} \parallel \overline{FG},
line segments DE and FG are parallel.\overline{KL} \perp \overline{DE},
line segments KL and DE are perpendicular.\green{\angle{GCJ}} = Tri_Y^\circ
\green{\angle{IAK}} = Tri_Y^\circ
What is
\blue{\angle{IAK}} = {?}
\blue{\angle{GCJ}} = {?}
NOTE: Angles not necessarily drawn to scale.
\Large{{}^\circ}
\orange{\angle{DAI}} = \green{\angle{GCJ}} = Tri_Y^\circ,
because they are alternate exterior angles, formed by two parallel lines
and a transversal line, they are congruent (equal).
label( [-.80, 2], "\\orange{" + Tri_Y + "^\\circ}",
"above left" );
arc( [0, 2], 1, 135, 180, {stroke: ORANGE} );
Alternatively, you can pair up using opposite angles and alternate interior
angles to achieve the same result (as shown in \pink{pink}).
label( [1, 2], "\\pink{" + Tri_Y + "^\\circ}",
"below right" );
arc( [0, 2], 1, 315, 360, {stroke: PINK} );
label( [3, -2], "\\pink{" + Tri_Y + "^\\circ}",
"above left" );
arc( [4, -2], 1, 135, 180, {stroke: PINK} );
\purple{\angle{DAK}} = 90^\circ,
because angles formed by perpendicular lines are equal to 90^\circ.
label( [-1.68, 2], "\\purple{90^\\circ}", "above left" );
arc( [0, 2], 1.65, 90, 180, {stroke: PURPLE} );
\blue{\angle{IAK}} = 90^\circ - \orange{\angle{DAI}} =
90 - Tri_Y^\circ,
because angles \blue{\angle{IAK}}
and \orange{\angle{DAI}} make up angle
\purple{\angle{DAK}}.
ORIGINAL_LABEL.remove();
label( [0, 3.5],
"\\blue{\\angle{IAK}} = " + (90 - Tri_Y) + "^\\circ",
"above left" );
\orange{\angle{DAI}} = 90^\circ - \green{\angle{IAK}} =
90 - Tri_Y^\circ,
because angles \green{\angle{IAK}}
and \orange{\angle{DAI}}, make up angle
\purple{\angle{DAK}}.
label( [-.80, 2], "\\orange{" + (90-Tri_Y) + "^\\circ}",
"above left" );
arc( [0, 2], 1, 135, 180, {stroke: ORANGE} );
\blue{\angle{GCJ}} = \orange{\angle{DAI}} =
90 - Tri_Y^\circ,
because they are alternate exterior angles formed by two parallel lines
and a transversal line, they are congruent (equal).
Alternatively, you can pair up using opposite angles and alternate interior
angles to achieve the same result (as shown in \pink{pink}).
label( [1, 2], "\\pink{" + (90-Tri_Y) + "^\\circ}",
"below right" );
arc( [0, 2], 1, 315, 360, {stroke: PINK} );
label( [3, -2], "\\pink{" + (90-Tri_Y) + "^\\circ}",
"above left" );
arc( [4, -2], 1, 135, 180, {stroke: PINK} );
ORIGINAL_LABEL.remove();
label( [4.75, -2],
"\\blue{\\angle{GCJ} = " + (90-Tri_Y) + "^\\circ}",
"below right" );