# Khan/khan-exercises

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 Inscribed angles 2
randRange( 0, 360 ) randRange( 10, 80 ) * 2 randRange( START + 180 + 5, START + CENTRAL + 180 - 5 ) % 360 4 CENTRAL "blue" "orange"

If the GIVEN_LABEL angle measures GIVEN degrees, what does the ASKED_LABEL angle measure?

This is a special case where the center of the circle is inside the inscribed orange angle. The blue angle is called a central angle.

init({ range: [ [ -RADIUS - 1, RADIUS + 1 ], [ -RADIUS - 1, RADIUS + 1 ] ] }); addMouseLayer(); graph.circle = new Circle( RADIUS ); style({ stroke: BLUE, fill: BLUE }); graph.circle.drawCenter(); graph.circle.drawPoint( START ); graph.circle.drawPoint( START + CENTRAL ); graph.central = graph.circle.drawCentralAngle( START, START + CENTRAL ); style({ stroke: ORANGE, fill: ORANGE }); graph.inscribed = graph.circle.drawInscribedAngle( SUBTENDED_POINT, START, START + CENTRAL ); graph.circle.drawMovablePoint( SUBTENDED_POINT, START + CENTRAL, START );
CENTRAL / 2 degrees

What do we know about the sub-angles formed by the dashed diameter shown above?

style({stroke: BLUE, "stroke-dasharray": "-"}, function() { graph.circle.drawChord( SUBTENDED_POINT, SUBTENDED_POINT + 180 ); });

If we only look at the sub-angles drawn now, we see that this is the special case from the previous inscribed angles exercise!

We know that the orange sub-angle is one half the measure of the blue sub-angle.

graph.inscribed.arc.animate({opacity: 0.4}); graph.central.arc.animate({opacity: 0.4}); style({stroke: BLUE}); graph.centralSub = graph.circle.drawCentralArc( SUBTENDED_POINT + 180, START + CENTRAL, 0.7 ); style({stroke: ORANGE}); graph.inscribedSub = graph.circle.drawInscribedArc( SUBTENDED_POINT, SUBTENDED_POINT + 180, START + CENTRAL, 0.7 );

Likewise, the other orange sub-angle is one half the measure of the other blue sub-angle, as shown.

graph.centralSub.remove(); graph.inscribedSub.remove(); style({stroke: BLUE}); graph.centralSub = graph.circle.drawCentralArc( START, SUBTENDED_POINT + 180, 0.7 ); style({stroke: ORANGE}); graph.inscribedSub = graph.circle.drawInscribedArc( SUBTENDED_POINT, START, SUBTENDED_POINT + 180, 0.7 );

If both orange sub-angles are one half both blue sub-angles, then we know that the original orange angle is one half the original blue angle.

graph.centralSub.remove(); graph.inscribedSub.remove(); graph.inscribed.arc.animate({opacity: 1.0}); graph.central.arc.animate({opacity: 1.0});

\color{ORANGE}{\text{orange angle}} = \dfrac{1}{2} \cdot \color{BLUE}{\text{blue angle}}

\color{ORANGE}{\text{orange angle}} = \dfrac{1}{2} \cdot \color{BLUE}{CENTRAL^{\circ}}

\color{ORANGE}{\text{orange angle}} = \color{ORANGE}{CENTRAL / 2^{\circ}}

CENTRAL / 2 "orange" "blue"
CENTRAL degrees

What do we know about the sub-angles formed by the dashed diameter shown above?

style({stroke: BLUE, "stroke-dasharray": "-"}, function() { graph.circle.drawChord( SUBTENDED_POINT, SUBTENDED_POINT + 180 ); });

If we only look at the sub-angles drawn now, we see that this is the special case from the previous inscribed angles exercise!

We know that the orange sub-angle is one half the measure of the blue sub-angle.

graph.inscribed.arc.animate({opacity: 0.4}); graph.central.arc.animate({opacity: 0.4}); style({stroke: BLUE}); graph.centralSub = graph.circle.drawCentralArc( SUBTENDED_POINT + 180, START + CENTRAL, 0.7 ); style({stroke: ORANGE}); graph.inscribedSub = graph.circle.drawInscribedArc( SUBTENDED_POINT, SUBTENDED_POINT + 180, START + CENTRAL, 0.7 );

Likewise, the other orange sub-angle is one half the measure of the other blue sub-angle, as shown.

graph.centralSub.remove(); graph.inscribedSub.remove(); style({stroke: BLUE}); graph.centralSub = graph.circle.drawCentralArc( START, SUBTENDED_POINT + 180, 0.7 ); style({stroke: ORANGE}); graph.inscribedSub = graph.circle.drawInscribedArc( SUBTENDED_POINT, START, SUBTENDED_POINT + 180, 0.7 );

If both blue sub-angles are twice both blue sub-angles, then we know that the blue is twice the orange angle.

graph.centralSub.remove(); graph.inscribedSub.remove(); graph.inscribed.arc.animate({opacity: 1.0}); graph.central.arc.animate({opacity: 1.0});

\color{BLUE}{\text{blue angle}} = 2 \cdot \color{ORANGE}{\text{orange angle}}

\color{BLUE}{\text{blue angle}} = 2 \cdot \color{ORANGE}{CENTRAL / 2^{\circ}}

\color{BLUE}{\text{blue angle}} = \color{BLUE}{CENTRAL^{\circ}}

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