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== 3.5#h(0.6em)Chapter 3 Activity


    #notebox("Activity", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Understanding the Law of Sines]


      - Materials: You will need paper and pencil, a ruler, compass, and protractor, or a dynamic geometry program such as Geogebra.


      + The vertices of a triangle determine a unique circumscribing circle. 
    + Draw a large acute triangle and label the vertices #math.equation(block: false, alt: "A")[$A$] , #math.equation(block: false, alt: "B")[$B$], and #math.equation(block: false, alt: "C")[$C$].
+ Draw the perpendicular bisector of side #math.equation(block: false, alt: "A C bar")[$limits(A C)^(―)$]. You can use a compass, or use your ruler and protractor as follows:+ Find the midpoint, #math.equation(block: false, alt: "M")[$M$], of side #math.equation(block: false, alt: "A C")[$A C$].
+ Draw a line that passes through #math.equation(block: false, alt: "M")[$M$] and is perpendicular to side #math.equation(block: false, alt: "A C bar")[$limits(A C)^(―)$].
+ Draw the perpendicular bisector of side #math.equation(block: false, alt: "B C bar")[$limits(B C)^(―)$] . Label the point where the two perpendicular bisectors meet with the letter #math.equation(block: false, alt: "O")[$O$].
+ From geometry, we know that every point on the perpendicular bisector of a segment #math.equation(block: false, alt: "P Q bar")[$limits(P Q)^(―)$] is equidistant between #math.equation(block: false, alt: "P")[$P$] and #math.equation(block: false, alt: "Q")[$Q$]. Use this fact to explain why the point #math.equation(block: false, alt: "O")[$O$] is the same distance from each vertex of your triangle.
+ Using the point at #math.equation(block: false, alt: "O")[$O$] as the center, draw the circle that passes through all three vertices of your triangle. The circle is called the #strong[circumscribing circle] of the triangle.

#figure(figph[triangle with circumscribing circle], alt: "triangle with circumscribing circle", caption: none)
+ We can move a vertex around the circle without changing the measure of its angle.
        + Measure angle #math.equation(block: false, alt: "C")[$C$] in your triangle. We say that this angle #strong[subtends] the arc joining #math.equation(block: false, alt: "A")[$A$] and #math.equation(block: false, alt: "B")[$B$] on the circle.
+ Choose any other point on the circle that is #emph[not] on the arc #math.equation(block: false, alt: "A B ⌢")[$limits(A B)^(⌢)$]. Call this point #math.equation(block: false, alt: "D")[$D$]. Draw the line segments #math.equation(block: false, alt: "A D bar")[$limits(A D)^(―)$] and #math.equation(block: false, alt: "B D bar")[$limits(B D)^(―)$] to create a second triangle. Measure the angle at #math.equation(block: false, alt: "D")[$D$]; it should be the same as the angle #math.equation(block: false, alt: "C")[$C$].
+ In fact, all angles with vertex on the circle that subtend the same arc have the same measure. Verify this fact by creating two more angles that subtend the arc #math.equation(block: false, alt: "A B ⌢")[$limits(A B)^(⌢)$].#figure(figph[triangles inscribed in circle], alt: "triangles inscribed in circle", caption: none)
+ What is the length of the diameter of the circumscribing circle?
        + At this point, if your drawing is too cluttered, you may need to make a fresh copy of the circumscribing circle. On this circle draw just the side #math.equation(block: false, alt: "A B bar")[$limits(A B)^(―)$] from your original triangle.
+ Draw the diameter that passes through #math.equation(block: false, alt: "B")[$B$] (remember that #math.equation(block: false, alt: "O")[$O$] is the center of the circle), and label the other end of the diameter with the letter #math.equation(block: false, alt: "C prime")[$C^(′)$].
+ Draw triangle #math.equation(block: false, alt: "Δ A B C prime")[$upright(Δ) A B C^(′)$], and measure the angle at #math.equation(block: false, alt: "C prime")[$C^(′)$]. It should be the same as angle #math.equation(block: false, alt: "C")[$C$] in your original triangle. (Why?) Let's call this angle #math.equation(block: false, alt: "θ")[$θ$].
+ Now measure angle #math.equation(block: false, alt: "∠ B A C prime")[$upright(∠) B A C^(′)$]. It should be 90 degrees, This is another fact from geometry: Any angle that intercepts the diameter of a circle is a right angle. Put a right angle symbol at angle #math.equation(block: false, alt: "∠ B A C prime")[$upright(∠) B A C^(′)$].
+ Label the length of the sides of your triangle as follows:
                - Side #math.equation(block: false, alt: "B C prime bar")[$limits(B C^(′))^(―)$] has length #math.equation(block: false, alt: "d")[$d$] (for diameter)
- Side #math.equation(block: false, alt: "A C prime bar")[$limits(A C^(′))^(―)$] has length #math.equation(block: false, alt: "b")[$b$]
- Side #math.equation(block: false, alt: "A B bar")[$limits(A B)^(―)$] has length #math.equation(block: false, alt: "c")[$c$]


                (Notice that #math.equation(block: false, alt: "c")[$c$] is the side opposite the angle #math.equation(block: false, alt: "θ")[$θ$]). Using the letters #math.equation(block: false, alt: "d")[$d$], #math.equation(block: false, alt: "b")[$b$], and #math.equation(block: false, alt: "c")[$c$] as needed, finish the equation
                    #math.equation(block: true, alt: "sin ⁡ θ equals")[$sin θ = #hide($000000000000000$)$]#figure(figph[circumscribed right triangle with sides labeled], alt: "circumscribed right triangle with sides labeled", caption: none)
+ Start from the last equation, and solve for #math.equation(block: false, alt: "d")[$d$] in terms of #math.equation(block: false, alt: "sin ⁡ C")[$sin C$] and #math.equation(block: false, alt: "c")[$c$].
+ Equating expressions for the diameter: The Law of Sines
            + You have now written an expression for the diameter of the circumscribing circle in terms of angle #math.equation(block: false, alt: "C")[$C$] and its opposite side. If you start the derivation with angle #math.equation(block: false, alt: "B")[$B$] and the arc #math.equation(block: false, alt: "A C ⌢")[$limits(A C)^(⌢)$], you will get another expression for the diameter. Write that expression here.
+ Now write an expression for the diameter that results from starting with angle #math.equation(block: false, alt: "A")[$A$] and arc #math.equation(block: false, alt: "B C ⌢")[$limits(B C)^(⌢)$].
+ Write an equation that reflects the fact that the diameter of the circle has the same length in all three expressions. #linebreak() This is the Law of Sines! All three expressions represent the diameter of the circumscribing circle, so they are equal to each other.
+ Verification for right and obtuse triangles #linebreak() If ABC is not an acute triangle, then one of the angles must either be a right angle or an obtuse angle.
            - In a right triangle, call the right angle #math.equation(block: false, alt: "A")[$A$]. Now you are already at step (4) of part C , so the derivation continues as before.

 #linebreak() To see the derivation for an obtuse triangle, follow the steps:
                + Draw a circle and a triangle with vertices #math.equation(block: false, alt: "A")[$A$], #math.equation(block: false, alt: "B")[$B$], and #math.equation(block: false, alt: "C")[$C$] on the circle so that there is an obtuse angle at point #math.equation(block: false, alt: "C")[$C$].
+ Measure the angle at #math.equation(block: false, alt: "C")[$C$]. and note that the larger arc #math.equation(block: false, alt: "A B ⌢")[$limits(A B)^(⌢)$] subtended by #math.equation(block: false, alt: "θ")[$θ$] is more than half of the circle.
+ Choose a point #math.equation(block: false, alt: "D")[$D$] on the larger arc #math.equation(block: false, alt: "A B ⌢")[$limits(A B)^(⌢)$]. In triangle #math.equation(block: false, alt: "Δ A B D")[$upright(Δ) A B D$], the angle at #math.equation(block: false, alt: "D")[$D$] subtends the shorter arc connecting #math.equation(block: false, alt: "A")[$A$] and #math.equation(block: false, alt: "B")[$B$]. Measure the angle at #math.equation(block: false, alt: "D")[$D$]. You should find that the angles at #math.equation(block: false, alt: "C")[$C$] and #math.equation(block: false, alt: "D")[$D$] are supplementary.
+ Here is another fact from geometry: The measure of an inscribed angle in a circle is half the measure of the arc it subtends. Use this fact to explain why the angles at #math.equation(block: false, alt: "C")[$C$] and #math.equation(block: false, alt: "D")[$D$] are supplementary.
+ What can you say about the sines of angles #math.equation(block: false, alt: "C")[$C$] and #math.equation(block: false, alt: "D")[$D$]? Because it is an acute triangle, we can use our original derivation on #math.equation(block: false, alt: "Δ A B D")[$upright(Δ) A B D$], and substitute #math.equation(block: false, alt: "C")[$C$] for #math.equation(block: false, alt: "D")[$D$] in the result.#figure(figph[obtuse circumscribed triangle], alt: "obtuse circumscribed triangle", caption: none)
]