Given the following:
\color{green}{\angle{ABC}} = Tri_Y°
\color{purple}{\angle{BAC}} = Tri_Z°
\overline{DE} \parallel \overline{BC}
What is \color{blue}{\angle{RAND_SWITCH2 === 0 ? "DAF" : "CAE"}} {?}
NOTE: Angles not drawn to scale.
Remember that the measure of the angles in a triangle sum to 180°.
Solve for \color{orange}{\angle{BCA}}
by subtracting the
measures of angles \color{purple}{\angle{BAC}}
and
\color{green}{\angle{ABC}}
from 180°. We find that
\color{orange}{\angle{BCA}} = Tri_X°
.
// label angle BAC
arc([2.5, 3], .75, 180, 220, { stroke: "orange" });
label([1.8, 3], "\\color{orange}{Tri_X°}",
"below left", {color: "orange"});
Solve for
\color{blue}{\angle{DAF}}
by using the fact
that it is a corresponding angle to
LABEL.remove();
LABEL = label([-3.3, 0],
"\\color{blue}{\\angle{DAF}}=Tri_X°",
"below");
\color{blue}{\angle{CAE}}
by using the fact
that it is an alternate interior angle to
LABEL.remove();
LABEL = label([1, 0],
"\\color{blue}{∠CAE} = Tri_X°",
"above", { color: "blue"});
\color{orange}{\angle{BCA}}.
That means those
angles are equal because they are both created by the same
set of parallel lines \overline{BC}
and \overline{DE}
, and transversal line
\overline{CF}
.
Given the following:
\overline{AB} \parallel \overline{CD}
(line AB is parallel to line CD)\color{purple}{\angle{EGB}} = X°.
\color{purple}{\angle{AGH}} = X°.
\color{purple}{\angle{BGH}} = 180 - X°
What is \color{blue}{\angle{RAND_SWITCH2 === 0 ? "GHD" : "CHF"}} {?}
NOTE: Angles not drawn to scale.
\color{blue}{\angle{GHD}} = \color{purple}{\angle{EGB}}
.
We know this because they are 2 complementary angles of a set of
parallel lines bisected by a single line.
First solve for \color{orange}{\angle{AGH}}
. We know that
\color{orange}{\angle{AGH}} = \color{purple}{\angle{EGB}}
because opposite angles are equal.
arc([1,2], .88, 180, 225, {stroke:"orange"});
label([0,2], "\\color{orange}{X°}", "below left");
\color{blue}{\angle{GHD}} = \color{purple}{\angle{AGH}}
We know this because they are 2 alternate interior angles of a set of
parallel lines bisected by a single line.
\color{blue}{\angle{CHF}} = \color{purple}{\angle{AGH}}
.
We know this because they are 2 corresponding angles of a set of
parallel lines bisected by a single line.
First solve for \color{orange}{\angle{AGH}}
. We know that
\color{orange}{\angle{AGH}} = 180° - \color{purple}{\angle{BGH}}
,
because angles along a line plane add up to 180°.
arc([1,2], .88, 180, 225, {stroke:"orange"});
label([0,2], "\\color{orange}{X°}", "below left");
\color{blue}{\angle{GHD}} = \color{orange}{\angle{AGH}}
.
We know those 2 angles are equal because they are alternate interior angles
of 2 parallel lines.
\color{blue}{\angle{CHF}} = \color{orange}{\angle{AGH}}
.
We know those 2 angles are equal because they are corresponding angles
formed by parallel lines, and a single bisecting lines.
Therefore,
\angle{GHD} = X°
LABEL.remove();
label([-2, -2], "\\color{blue}{\\angle{GHD}}=X°",
"above right");
\angle{CHF} = X°
LABEL.remove();
label([-4, -2.5], "\\color{blue}{\\angle{CHF}}=X°",
"below left");
.
Given the following:
\color{green}{\angle{BDC}°} = Tri1_Y
\color{orange}{\angle{DBE}°} = Tri1_X
What is \color{blue}{\angle{RAND_SWITCH3 === 0 ? "CHE" : ( RAND_SWITCH3 === 1 ? "GHC" : "DHE" )}} {?}
NOTE: Angles not drawn to scale.
Tri1_Z 180-Tri1_Z
\color{purple}{\angle{BHD}} =
180° - \color{green}{\angle{BDC}} - \color{orange}{\angle{DBE}} =
Tri1_Z°.
This is because the interior angles of a triangle add up to 180°.
// label angle BHD
arc([3.2, 1.3], .75, 118, 220, { stroke: "purple" });
label([2.6, 2], "\\color{purple}{Tri1_Z^\\circ}",
"below left");
\color{blue}{\angle{CHE}} = \color{purple}{\angle{BHD}}
.
This is because they are opposite each other, and opposite angles are
equal.
\color{blue}{\angle{CHE}} = Tri1_Z°
\color{blue}{\angle{CHG}}
\color{blue}{\angle{DHE}}
= 180° - \color{purple}{\angle{BHD}}
.
This is because angles along a line add up to
180°.
\color{blue}{\angle{GHC}}
LABEL.remove();
label([4, 2.5],
"\\color{blue}{\\angle{GHC}}=180 - Tri1_Z^\\circ",
"above");
\color{blue}{\angle{DHE}}
LABEL.remove();
label([2.5, -0.5],
"\\color{blue}{\\angle{DHE}}=180 - Tri1_Z^\\circ",
"below");
= 180 - Tri1_Z°
Given the following:
\color{green}{\angle{BDC}°} = Tri2_Y°
\color{orange}{\angle{AIC}°} = 180 - Tri2_Z°
\color{green}{\angle{GCH}°} = Tri2_Y°
\color{orange}{\angle{FGH}°} = 180 - Tri2_Z°
What is \color{blue}{\angle{RAND_SWITCH2 === 0 ? "AJF" : "IHE"}} {?}
NOTE: Angles not drawn to scale.
\color{purple}{\angle{DIJ}} = 180° - \color{orange}{\angle{AIC}}
.
This is because angles along a line total 180°.
// label angle JID
arc([0, -1.2], .75, 143, 220, { stroke: "purple" });
label([-.75, -1.2], "\\color{purple}{Tri2_Z°}",
"left");
\color{purple}{\angle{HGC}} = 180° - \color{orange}{\angle{FGH}}
.
This is because angles along a line or flat plane total 180°.
// label angle HGC
arc([1.8, 5], 1, 280, 0, { stroke: "purple" });
label([2.5, 4.3], "\\color{purple}{Tri2_Z°}",
"below right");
\color{teal}{\angle{DJI}} = 180° - \color{green}{\angle{BDC}} -
\color{purple}{\angle{DIJ}}
.
We know this because the sum of angles inside a triangle add up to 180°.
// label angle JID
arc([-3.2, 1.3], .75, 260, 320, { stroke: "teal" });
label([-3.2, 0.50], "\\color{teal}{Tri2_X°}",
"below right");
\color{teal}{\angle{CHG}} = 180° - \color{green}{\angle{ACD}} -
\color{purple}{\angle{HGC}}
.
We know this, because the sum of angles inside a triangle add up to 180°.
// label angle CHG
arc([3.2, 1.3], .75, 38, 120, { stroke: "teal" });
label([3.4, 1.78], "\\color{teal}{Tri2_X°}",
"above");
\color{blue}{\angle{AJF}} = \color{teal}{\angle{DJI}}
.
We know they are equal because they are opposite angles.
\color{blue}{\angle{IHE}} = \color{teal}{\angle{CHG}}
.
We know they are equal because they are opposite angles.
Therefore,
\angle{AJF} = Tri2_X°
LABEL.remove();
label([-3.7, 2.5],
"\\color{blue}{\\angle{AJF}}=Tri2_X°", "above");
\angle{IHE} = Tri2_X°
LABEL.remove();
label([4.0, -0.3],
"\\color{blue}{\\angle{IHE}}=Tri2_X°",
"below left");
.