1.
Area Formula
(a) Consider figure 1 below, by checking with gsp the area of triangle ABC is 2.92 inches. One the other hand, using the area formula it also gives the area to be 2.92 inches. It seems the formula holds in finding the area of the triangle. To see that the area formula holds for all cases, click here to play with the gsp file.
(b) Proof
Case 1 (angle C is acute)
Let ABC be a triangle drop the perpendicular (Existence of perpendicular theorem) from the point B to the line AC and let D be the foot of the perpendicular as shown below.Then from definition, the sine of the angle ACB is the ratio BD/DC, where angle ABC = C, BD = h and BC = a. Therefore sin(C) = h/a so h = a*sin(C). We know that area of a triangle is 1/2*b*h, but in this case h = a*sin(C). By substituting h for a*sin(C) in the area of the triangle i.e 1/2*b*h, we get Area = 1/2*a*b*sin(C)
Case 2 (angle C is obtuse)
Let ABC be a triangle drop the perpendicular(Existence of perpendicular theorem) from the point B to the line AC and let D be the foot of the perpendicular as shown below.Then from definition, the sine of the angle ACB is the ratio BD/DC, where angle ACB = C, BD = h and BC = a. In this case, to find h consider triangle BCD. Now since we are using triangle CBD, the the angle to consider is angle BCD. From straight angle theorem, angle ACB and BCD are supplementary. Therefore angle BCD is (180 - C) but from definition sin(C) is equal to sin(180-C). We know that area of a triangle is 1/2*b*h, but in this case h = a*sin(180 - C) = a*sin(C) as defined from definition. By substituting h for a*sin(C) in the area of the triangle i.e 1/2*b*h, we get: Area = 1/2*a*b*sin(C)
(c) Application
The area formula can be used in several places. For example finding the area of any a polygon. For example in the figure ABCDE below, the area can easily be determined easily by dividing the polygon into triangle of any size. By using two of the sides and and an angle between them, we can determine the area of each triangle easily by using the area formula. Finally take each area of the triangle and sum them up. It should be noted that for area formula to be used, two sided of the triangle and angle between them has to be known.
Law of cosines
(a) Consider figure 1 constructed using gsp by letting AC = c, AC = b and BC = a. It seems that as we take any value of c and find its square, the result will be equal to a^2 + b^2 - 2*a*b*cos(C). To verify that it seems true for all cases click here.
(b) Proof
case 1 Angle C is acute
Let BD be the altitude of the triangle from the vertex B. let h = BD, x = AD, and y = DC. By definition of cosine, cos(C) = y/a, so a*b*cos(C) = a*b*(y/a) = by. By Pythagorean theorem, c^2 = x^2 + h^2 and a^2 = y^2 + h^2. Eliminating h from these two equations, we get c^2 = x^2 + a^2 - y^2.. (eq 1). Now substitute b-y for x in equation 1.
c^2 = (b-y)^2 + a^2 - y^2
c^2 = b^2 - 2by + y^2 + a^2 - y^2
c^2 = a^2 + b^2- 2by, but by = abcosC
c^2 = a^2 + b^2 - 2abcosC.
Case 2 Angle C is a right angle
If angle C is a right angle then cosC = 0 therefore the law of cosine is Pythagorean theorem i.e c^2 =a^2 + b^2
Case 3 Angle C is obtuse
With the same approach, Let BD be the altitude of the triangle from the vertex B. let h = BD, x = AD, and y = DC. By definition of cosine, cos(180-C) = y/a, so a*b*cos(180- C) = a*b*(y/a) = by. By Pythagorean theorem, c^2 = (x + y)^2 + h^2 and a^2 = y^2 + h^2. Eliminating h from these two equations, we get c^2 = (x+y)^2 + a^2 - y^2.. (eq 1). Now substitute b+y for x in equation 1.
c^2 = (b+y)^2 + a^2 - y^2
c^2 = b^2 + 2by + y^2 + a^2 - y^2
c^2 = a^2 + b^2+ 2by, but by = abcos(180-C) = -abcos C (by definition) c^2 = a^2 + b^2 - 2abcos C.
(C) Application
The law of cosine is very
useful in determining angles of any triangular figure as long
as the three sides of the figure are known. Also it can be used
to determine one of the sides of the triangle when one angle and
two sides are given.
Sine rule
(a)
It appears from the gsp image below that the sine rule hold. That is to say a/sinA, b/sinB and c/sinC gives the same value i.e 4.16 as shown below. To show that it seems to hold for all cases, click here to play the gsp file.
(b) Proof
Let h be the altitude of
triangle ABC from the vertex B Then, by definition of the sine
function, sin A = h/c and sin C = h/a. Thus csinA = h and a sin
C = h. Therefore c sin A = a sin C, so c/sinC = a/sin A.
Similarly, let k be the
altitude of triangle ABC from the vertex A. Then sin B = k/c and
sin C = k/b, so c/sinC = b/sin B . Therefore we can conclude that
a/sin A = b/sin B = c/sin C
(c) Application
The sine rule is applied
when two sides and one angle or one angle and two sides are known.
With either of the two conditions satisfied, the other measurement
of the triangle can easily be determined.
2.
Proof that for a triangle ABC labeled as in problem 1, with circumradius R, the area of ABC is equal to abc/4R.
It can be noted from the image below that using the formula and gsp to find area, both gives the same area as a*b*c/4R. That is the conjecture I will like to prove if it is always true. As we noted in problem 1, the area of triangle ABC is 1/2absinC. So we need to prove that sinC = c/2R
case 1: angel C is acute.
Let the circumcircle of triangle ABC be c1 with diameter AD = 2R. Using the inscribed angle theorem, angle BCA is equal to angle BDA. Also by the same theorem, triangle ABD is right with 2R as the hypotenuse. By definition, sine (BDA) = c/2R from the inscribed angle theorem, sin(C) = sine (BDA) = c/2R. Now given the area formula of triangle ABC is 1/2*a*b*sin(C) and sin (C) = sin (BDA) = c/2R, then the area of triangle ABC will be 1/2*a*b*sin (C) = 1/2*a*b*sin(BDA) = 1/2*a*b*c/2R =
Case 2: Angle C is obtuse
Let the circumcircle of triangle ABC be c1 with diameter AD = 2R. Using the inscribed angle theorem, angle (180 - BCA) is equal to angle BDA (note that ACBD is cyclic quadrilateral i.e angle D = angle (180 - C)). Also by the same theorem, triangle ABD is right with 2R as the hypotenuse. By definition, sine (BDA) = c/2R from the inscribed angle theorem, sin(180 - C) = sin(C) = sine (BDA) = c/2R. Now given the area formula of triangle ABC is 1/2*a*b*sin(C) and sin (C) = sin (BDA) = c/2R, then the area of triangle ABC will be 1/2*a*b*sin (C) = 1/2*a*b*sin(BDA) = 1/2*a*b*c/2R =
To play with the gsp file that created all this image, click here
3.
Let ABC be a triangle with circumradius R. Suppose that C1 and C2 are circles such that both C1 and C2 pass through the point A, C1 is tangent to BC at B, and C2 is tangent to BC at C. Let p be the radius of C1 and let q be the radius of C2. Prove that pq = R^2.
Construction:
Construct triangle ABC and using its sides construct circumcircle passing through vertex A,B and C. Now given circle c1 pass through the point A and is tangent to BC at B, by definition the radius of c1 is tangent to BC. Therefore construct a line which is perpendicular to BC at B and given the circle pass through vertex A then it means that from the center of c1 to A is the radius of c1. Construct the midpoint of AB and from it construct a perpendicular line to meet line perpendicular to BC at B at H. Using H and A construct the segment which is the radius of c1. Now given HB = HA, then HB = HA = p. Finally use p as the radius and H as the center of circle c1 to construct c1. With the same argument, constuct q and center F to construct c2. From the gsp file it shows that p*q = R^2 = 17.23.9 click here to play with the gsp file.
Proof 
In the figure above let H be the centre of c1, let F be the center of c2 and O be the center of c3. Note we have created five triangles, AHB, ABO, OBC, OAC and AFC which are all isoceles i.e by using isoceles triangle theorem, each of these triangles has its base angles equal. Let angle HAB = HBA = x, OAB = OBA = y, OAC = OCA = z, OCB = OBC = m, and FAC = FCA = r. Given HB is perpendicular to BC at B, then angle HBC = x + y + m = 90 therefore
(i) x + y + m = 90
With the same point, FC is tangent to line BC at C hence angle FCB = r + z + m = 90, therefore
(ii) r + z + m = 90
Finally by using triangle sum theorem, of triagle ABC, then (y + z) + (y + m) + (z + m) = 180 i.e 2y + 2z + 2m = 280 by simplifying we get
(iii) z + y + m = 90
Solving (i) and (iii) we get z = x also solving (ii) and (iii) we get y = r.
By angle angle similarity, triangle ABO is similar to ACF and triangle COA is similar to triangle AHB.
Given triangle ABO is similar to ACF, then
R/c = q/b which means q = Rb/c ...(1)
Also given COA is similar to AHB, then R/b = p/c which means p = Rc/b ...(2)
Therefore, pq = Rc/b*Rb/c = R^2
qed.
4.
Prove that for a triangle ABC labelled as in problem 1, if D is the midpoint of side AB, then the length m of the median CD satisfies the equation m^2 = 1/2 a^2 + 1/2 b^2 - 1/4 c^2.
From the gsp file which created the image below, it shows that the median CD satisfies the equation m^2 = 1/2a^2 + 1/2b^2 - 1/4c^2 which need to be proved. Click here to see the gsp file
Consider triangle DBC, using the law of cosine to angle B gives
m^2 = a^2 + (1/2c)^2 - 2a(1/2c)cosB, i.e
m^2 = a^2 + 1/4c^2 - a*c*cos(B) ...(i)
Cosider triangle ABC using the law of cosine to angle B gives
b^2 = a^2 + c^2 - 2*a*c*cos(B), so
ac cosB = 1/2(a^2 + c^2 - b^2). ...(ii)
By substituting (ii) in (i) we get m^2 = a^2 + 1/4c^2 - 1/2( a^2 + c^2 - b^2),
m^2 = a^2 + 1/4c^2 - 1/2 a^2 - 1/2c^2 + 1/2b^2
m^2 = 1/2 a^2 + 1/2b^2 - 1/4c^2.
qed
