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Mathematics Dictionary
Dr. K. G. Shih
Angles and sides of triangle
Subjects


  • TR 10 00 | - Outlines
  • TR 10 01 | - Sine law
  • TR 10 02 | - Cosine law
  • TR 10 03 | - Solve triangle if SSA or SAA are given
  • TR 10 04 | - Solve triangle if SAS are given
  • TR 10 05 | - Sin(A) = 2*Sqr(s*(s-a)*(s-b)*(s-c))/(b*c)
  • TR 10 06 | - Area of triangle = Sqr(s(s-a)*(s-b)*(s-c)/(b*c))
  • TR 10 07 | - tan(A/2) = Sqr((s-b)*(s-c)/(s*(s-a))
  • TR 10 08 | - cos(A/2) = Sqr(s*(s-a)/(b*c))
  • TR 10 09 | - sin(A/2) = Sqr((s-b)*(s-c)/(b*c)
  • TR 10 10 | - In-circle
  • TR 10 11 | - Es-circle
  • TR 10 12 | - tan(A/2)*tan(B/2) + tan(B/2)*tan(C/2) + tan(C/2)*tan(A/2) = 1
  • TR 10 13 | - Circum-circle
  • TR 10 14 | - Triangle ABC by joining 3 points E,F,T on sides and find area EFT
  • TR 10 15 | - Triangle ABC by joining 3 points E,F,T on sides and find area EFT
  • TR 10 16 | - Triangle ABC by joining 3 points E,F,T on sides and find area EFT
  • TR 10 17 | - Triangle ABC by joining 3 points E,F,T on sides and find area EFT
    • Answers


    TR 10 01. Sine law

    • Sine law.
      • Triangle ABC
        • Angles A,B and C .
        • sides a = BC, b = CA and c = AB.
      • Defintion of sine law
        • Triangle ABC : a/sin(A) = b/sin(B) = c/sin(C) = 2*R.
        • Where R is theradius of circum-circle.
      • Application : Solve triangle to find another angles and sides
        • SAA : Solve triangle if a, A, B are given.
        • SSA : Solve triangle if a, b, A are given.
    • Method 1 : Use right triangle inscribed a circle
    • Construction
      • Draw a circle of radius R.
      • Draw triangle inscribed in the circle.
      • Draw diameter BP and formed triangle PBC.
    • Proof
      • Angle BPC = 90 degrees.
      • Hence sin(BPC) = BC/BP = a/(2*R)
      • Angle BPC = angle BAC = A.
      • Hence sin(A) = sin(BPC) = a/(2*R)
      • Hence a/Sin(A) = 2*R.
      • Similarly, b/sin(B) = 2*R and c/sin(C) = 2*R
  • Method 2 : Use use triangle area = b*c*sin(A)/2
    • Triangle Area = b*c*sin(A) = c*a*sin(B) = a*b*sin(C).
    • Divide by abc.
    • Hence sin(A)/a = sin(B)/c = sin(C)/c.

    Go to Begin

    TR 10 02. Cosine Law

    • Cosine law.
      • Defintion : Trianle ABC
        • a^2 = b^2 + c^2 - 2*b*c*cos(A).
        • b^2 = c^2 + a^2 - 2*c*a*cos(B).
        • c^2 = a^2 + b^2 - 2*b*c*cos(C).
      • Application : Solve triangle if SAS are given
        • Solve triangle if b, A, c are given.
        • Solve triangle if c, B, a are given.
        • Solve triangle if a, C, b are given.
    • Proof of cosine law
    • Construction
      • Draw triangle ABC.
      • Draw AH perpendicular to AB and let AH = h.
      • Let AH = x and BH = c - x
    • Proof of a^2 = b^2 + b^2 - 2*b*b*cos(A).
      • Triangle BCH
        • a^2 = h^2 + (c - x)^2.
        • a^2 = h^2 + c^2 + x^2 - 2*c*x.
      • Triangle AHC : x = b*cos(A) and h = b*sin(A)
      • a^2 = (b^2)*sin(A)^2 + b^2)*cos(A)^2 + c^2 - 2*b*c*cos(A).
      • Since sin(A)^2 + cos(A)^2 = 1
      • Hence a^2 = b^2 + c^2 - 2*b*c*cos(A)
    • Application : Sum of two vectors
      • From A walks 2 km to east at B. Then walks 1 km to N.E. at C.
      • Find the distance and direction between A and C.
      • Solution
        • let AC = b, AB = c = 2 km, BC = a = 1 km.
        • Angle ABC = B = 180 - 45 degrees = 135 degrees.
        • Hence b^2 = a^2 + c^2 - 2*a*c*cos(135)
        • Hence b^2 = 1^2 + 2^2 + 4*cos(45) = 5 + 2*Sqr(2).
        • Hence b = Sqr(5 + 2*Sqr(2)) = Sqr(7.828427) = 2.79793 km.
        • Find direction by sine law in triangle ABC.
        • a/sin(A) = b/sin(B)
        • Hence sin(A) = (a*sin(B))/b = sin(135)/2.79793 = 0.50545.
        • Hence A = arcsin(0.50545)30.3612 degrees.
    • Application : Use cosine law to prove
      • sin(A/2) = Sqr((s-b)*(s-c)/(b*c)).
      • cos(A/2) = Sqr(s*(s-a)/(b*c)).

    Go to Begin

    TR 10 03. Solve triangle if SSA or SAA are given

    Given that A = 30, a = 4 and b = 6. Find B, C and c.
    • Using sine law
    • Sin(A)/a = sin(B)/b
    • Hence sin(B) = b*sin(A)/a = 6 *sin(30)/4 = 1.5*(1/2) = 0.75.
    • Hence B = 48.59 degrees.
    • C = 180 - A - B = 101.41.
    • Sin(A)/a = sin(C)/c
    • c = a*sin(C)/sin(A) = 4*sin(101.41)/sin(30) = 4*(0.980237)/(1/2) = 7.8419.
    Verify
    • sin(A)/a = sin(30)/4 = 1/8 = 0.125.
    • sin(C)/c = sin(101.41)/7.6419 = 0.12499

    Go to Begin

    TR 10 04. Solve triangle if SAS are given
    Example : if a = 2 and b = 3 and angle C = 60 degrees, find other side c.
    • c^2 = 2^2 + 3^2 - 2*2*3*cos(60).
    • c^2 = 4 + 9 - 12*(1/2).
    • c^2 = 7.0
    • Hence c = 2.64575.

    Go to Begin

    Q05. Sin(A) = 2*Sqr(s*(s-a)*(s-b)*(s-c))/(b*c)

    Method 1
    • Hint : Use cosine law a^2 = b^2 + c^2 - 2*b*c*cos(A)
    Proof
    • cos(A) = (b^2 + c^2 - a^2)/(2*b*c).
    • sin(A)^2 = 1 - cos(A)^2.
    • sin(A)^2 = 1 - (b^2 +c^2 - a^2)^2/(4*b^2*c^2).
    • sin(A)^2 = (1 - (b^2 +c^2 - a^2)/(2*b*c))*(1 + (b^2 +c^2 - a^2)/(2*b*c)).
    • sin(A)^2 = ((b+c)^2 - a^2)*(a^2 - (b+c)^2)/(4*b^2*c^2).
    • sin(A)^2 = (a+b+c)*(-a+b+c)*(a-b+c)*(a+b-c)/(4*b^2*c^2).
    • Let s = (a+b+c)/2.
    • Hence a+b+c = 2*s, -a+b+c = 2*(s-a), a-b+c = 2*(s-b0 and a+b-c = 2*(s-c)
    • Hence sin(A)^2 = 16*s*(s-a)(s-b)*(s-c)/(4*b^2*c^2)
    • Hence sin(A)^2 = 4*s*(s-a)*(s-b)*(s-c)/(b^2*c^2)
    • Hence sin(A) = 2*Sqr(s(s-a)*(s-b)*(s-c)/(b*c))/(b*c).
    Method 2
    • Hint : Use sin(A/2) = Sqr((s-b)*(s-c)/(b*c)).
    • Use cos(A/2) = Sqr(s*(s-a)/(b*c)).
    • Use sin(A) = 2*sin(A/2)*cos(A/2).
    Proof
    • Hence sin(A) = 2*Sqr(s*(s-a)(s-b)*(s-c))/(b*c).

    Go to Begin

    TR 10 06. Area of triangle = Sqr(s(s-a)*(s-b)*(s-c)/(b*c))

    Proof
    • Since area of triangle = b*c*sin(A)/2
    • Hence area of triangle = (b*c)*2*Sqr(s(s-a)*(s-b)*(s-c)/(b*c))/(b*c).
    • Hence area of triangle = Sqr(s(s-a)*(s-b)*(s-c)/(b*c)).
    • This is called Heron formula (MD 2002 program 20 13 or 20 16).

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    TR 10 07. tan(A/2) = Sqr((s-b)*(s-c)/(s*(s-a))

    Method 1
    • Hint : Use Area of triangle = r*s where r is radius of incircle.
    • Use r = (s-a)*tan(A/2).
    Proof
    • tan(A/2) = r/(s-a).
    • r = Area/s.
    • r = Sqr(s*(s-a)*(s-b)*(s-c))/s.
    • Hence tan(A/2) = Sqr((s-b)*(s-c)/(s*(s-a)).
    Method 2
    • Hint : Hence tan(A/2) = Sqr((s-b)*(s-c)/(s*(s-a)).
    Identities
    • tan(A/2) = Sqr((s-b)*(s-c)/(s*(s-a)).
    • tan(B/2) = Sqr((s-c)*(s-a)/(s*(s-b)).
    • tan(C/2) = Sqr((s-a)*(s-b)/(s*(s-c)).

    Go to Begin

    TR 10 08. cos(A/2) = Sqr(s*(s-a)/(b*c))

    Method 1
    Hint
    • 1. Use 1 + tan(x)^2 = sec(x)^2.
    • 2. Use cos(x) = 1/sec(x).
    Proof
    • cos(A/2)^2 = 1/(Sec(A/2)^2).
    • cos(A/2)^2 = 1/(1 + tan(A/2)^2).
    • cos(A/2)^2 = 1/(1 + (s-b)*(s-c)/(s*(s-a))
    • cos(A/2)^2 = s*(s-a)/(s*(s-a) + (s-b)*(s-c)).
    • s*(s-a) + (s-b)*(s-c) = s^2 - s*a + s^2 - s*(b+c) + b*c.
    • s*(s-a) + (s-b)*(s-c) = 2*s^2 - s*(a+b+c) + b*c.
    • s*(s-a) + (s-b)*(s-c) = 2*s^2 - s*(2*s) + bc = b*c.
    • Hence cos(A/2) = Sqr(s*(s-a)/(b*c)).
    Method 2
    Hint
    • 1. Use cosine law.
    • 2. Use cos(2*x) = 2*cos(x)^2 - 1.
    • 3. Use a^2 + 2*b*c + c^2 = (b+c)^2.
    Proof
    • Cosine law : cos(A) = (b^2 + c^2 - a^2)/(b*c).
    • Since cos(2*x) = 2*cos(x)^2 -1.
    • Hence cos(A/2)^2 = (1 + cos(A))/2.
    • Hence cos(A/2) = Sqr(1 + (b^2 + c^2 - a^2)/(2*b*c)).
    • Hence cos(A/2) = Sqr(2*b*c + (b^2 + c^2 - a^2))/(2*b*c)).
    • Hence cos(A/2) = Sqr((b + c)^2 - a^2)/(2*b*c)).
    • Hence cos(A/2) = Sqr((b + c + a)*(b + c - a)/(2*b*c)).
    • Let 2*s = a + b + c and then b + c - a = a + b + c - 2*a = 2*s - 2*a.
    • Hence cos(A/2) = Sqr((2*s)*(2*s - 2*a)/(2*b*c)).
    • Hence cos(A/2) = Sqr(s*(s - a)/(b*c)).
    Identities
    • cos(A/2) = Sqr(s*(s-a)/(b*c)).
    • cos(B/2) = Sqr(s*(s-b)/(c*a)).
    • cos(C/2) = Sqr(s*(s-c)/(a*b)).

    Go to Begin

    TR 10 09. sin(A/2) = Sqr((s-b)*(s-c)/(b*c)

    Method 1
    Hint
    • Use sin(x) = tan(x)*cos(x).
    • tan(A/2) = Sqr((s-b)*(s-c)?(s*(s-a)).
    • cos(A/2) = Sqr(s*(s-a)/(b*c)).
    Proof
    • sin(A/2)^2 = tan(A/2)*cos(A/2).
    • sis(A/2)^2 = Sqr((s-b)*(s-c)/(s*(s-a))*Sqr(s*(s-a)/(b*c)).
    • sin(A/2)^2 = Sqr((s-b)*(s-c)/(b*c)).
    • Hence sin(A/2) = Sqr((s-b)*(s-c)/(b*c)).
    Method 2
    Hint
    • 1. Use cosine law.
    • 2. Use cos(2*x) = 1 - 2*sin(x)^2.
    • 3. Use a^2 - 2*b*c + c^2 = (b-c)^2.
    Proof
    • Cosine law : cos(A) = (b^2 + c^2 - a^2)/(b*c).
    • Since cos(2*x) = 1 - 2*sin(x)^2.
    • Hence sin(A/2)^2 = (1 - cos(A))/2.
    • Hence sin(A/2) = Sqr(1 - (b^2 + c^2 - a^2)/(2*b*c)).
    • Hence sin(A/2) = Sqr(2*b*c - (b^2 + c^2 - a^2))/(2*b*c)).
    • Hence sin(A/2) = Sqr(a^2 - (b - c)^2 )/(2*b*c)).
    • Hence sin(A/2) = Sqr((a + b - c)*(a + c - b)/(2*b*c)).
    • Let 2*s = a + b + c and then a + b - c = a + b + c - 2*c = 2*s - 2*c.
    • Hence sin(A/2) = Sqr((2*s)*(2*s - 2*a)/(2*b*c)).
    • Hence sin(A/2) = Sqr((s - b)*(s - c)/(b*c)).
    Identities
    • sin(A/2) = Sqr((s-b)*(s-c)/(b*c)).
    • sin(B/2) = Sqr((s-c)*(s-a)/(c*a)).
    • sin(C/2) = Sqr((s-a)*(s-b)/(a*b)).

    Go to Begin

    TR 10 10. In-circle

    Defintion
    • Bisectors of internal angles of triangle are concurrent which is in-center.
    • In-radius is radius of in-circle which is r.
    Construction
    • draw a traingle ABC.
    • Draw an in-circle with center I.
    • The incircle tangents the sides of triangle ABC.
    • Let a,b,c are sides and 2*s = a + b + c.
    AF = s - a : Length of tangent at A to incircle.
    • Let incircle tangent AB at F, tangent BC at D and tangent AC at E.
    • Hence tangent at A is AE and AF and AE = AF.
    • Hence tangent at B is BD and BF and BD = BF.
    • Hence tangent at C is CD and CE and CD = CE.
    • AE + AF = (b + c) - (CE + BF).
    • CE + BF = BD + CD = a
    • Hence 2*AE = (b + c) - a = (a + b + c) - 2*a. = 2*s - 2*a
    • Hence AF = AE = s - a.
    • Similarly BF = s - b. and CD = s - c.
    Show that area of triangle = r*s.
    • Area IAB = IF*c/2 = r*c/2.
    • Area IBC = ID*a/2 = r*a/2.
    • Area ICA = IE*b/2 = r*b/2.
    • Area ABC = IAB + IBC + ICA = r*(a + b + c)/2 = r*s.

    Go to Begin

    TR 10 11. Es-circle

    Defintion
    • Bisectors of one internal and two external angles of triangle are concurrent.
    • The point is Es-center.
    • Triangle has 3 es-circles.
    • Es-radius is radius of es-circle which is r1, r2 and r3.
    Cconstruction
    • draw a traingle ABC.
    • Draw an es-circle with center P.
    • The es-circle tangents the sides of triangle ABC.
    • Let a,b,c are sides and 2*s = a + b + c.
    AF = s : the length of tangent at A to es-circle.
    • Let es-circle tangent AB at F, tangent BC at D and tangent AC at E.
    • Hence tangent at A is AE and AF and AE = AF.
    • Hence tangent at B is BD and BF and BD = BF.
    • Hence tangent at C is CD and CE and CD = CE.
    • AE + AF = (b + c) + (CE + BF).
    • CE + BF = BD + CD = a
    • Hence 2*AE = (b + c) + a = s.
    • Hence AF = AE = s.
    Prove that area of triangle = r1*(s-a) = r2*(s-b) = r3*(s-c)
    • Let I be the es-center.
    • Area of triangle ABC = area ABIC - area BIC
    • Area ABIC = area ABI + area ACI = r1*c/2 + r1*b/2
    • Area BIC = r1*a/2
    • Hence area ABC = r1*(b + c - a)/2 = r1*(a + b + c - 2*a)/2 = r1 * (s-a)

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    TR 10 12. tan(A/2)*tan(B/2) + tan(B/2)*tan(C/2) + tan(C/2)*tan(A/2) = 1

    Hint
    • Use tan(A/2) = Sqr((s-b)*(s-c)/(s*(s-a)).
    • s = a + b + c and A, B, C are angle of triangle.
    Proof
    • tan(A/2)*tan(B/2) = Sqr((s-b)*(s-c)/(s*(s-a))*Sqr((s-c)*(s-a)/(s*(s-b)).
    • tan(A/2)*tan(B/2) = Sqr((s-c)*(s-c)/(s*s) = (s-c)/s.
    • Similarly tan(B/2)*tan(C/2) = (s-a)/s.
    • Similarly tan(C/2)*tan(A/2) = (s-b)/s.
    • Hence LHS = (s-a)/s + (s-b)/s + (s-c)/s = (3*s - a - b - c)/s = s/s = 1.

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    TR 10 13. Circum-circle

    Defintion
    • Bisectors of sides of triangle are concurrent which is Circum-center.
    • It is also called as circum-center.
    • Circum-radius is radius of circum-circle which is R.
    Cconstruction
    • draw a traingle ABC.
    • Draw an circum-circle with center E.
    • The circum-center has same distance from vertices A, B, C.
    • The circum-radius is R.
    Prove that area of triangle = abc/(4*R)
    • Since a = 2*R*sin(A).
    • Area ABC = b*c*sin(A)/2.
    • Area ABC = (b*c)*(a/(2*R))/2.
    • Area ABC = abc/(4*R).

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    TR 10 14. Four tringales inside triangle ABC

    Construction
    • Draw a triangle ABC and sides are a,b,c
    • Let E be a point on AC and CE = 2*b/3 and AE = b/3
    • Let T be a point on BA and BA = c
    • Let F be a mid point of BC and BC = a
    • Join ET, TF and FE so that make four triangles
    • Let area of triangle BTF = x
    • Let area of triangle EFC = y
    • Let area of triangle ATE = z
    Question
    • If x^2 = y*z, find the ratio BT : TA
    • Find the area of triangle ETF
    Solution
    • Keyword | Find area BT : AT if BF = BC/2, CE = 2*CA/3 and (BTF)^2 = (CEF)*(AET)

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    TR 10 15. Divide tringale ABC into 4 equal area triangles

    Construction
    • Draw a triangle ABC and sides are a,b,c
    • Let E be mid point on AC and CE = AE = b/2
    • Let T be mid point on BA and BT = TA = c/2
    • Let F be mid point of BC and BF = CF = a/2
    • Join ET, TF and FE so that make four triangles
    Question
    • Prove that area FEC = area EAT = area TBF = area EFT
    • Area FEC = (R^2)*sin(A)*sin(B)*sin(C)/2
    Solution
    Go to Begin

    TR 10 16. Four tringales inside triangle ABC by joining E,F,T on sides

    Construction
    • Draw a triangle ABC and sides are a,b,c
    • Let E be a point on AC and EA = b/n
    • Let T be a point on BA and BT = c/n
    • Let F be a point on BC and FB = a/n
    • Join ET, TF and FE so that make four triangles
    Question
    • Find the area of triangle ETF = (area ABC)*(1 - 3*(n-1))/(n^2)
    Solution
    Go to Begin

    TR 10 17. Four tringales inside triangle ABC by joining E,F,T on sides

    Construction
    • Draw a triangle ABC and sides are a,b,c
    • Let E be a point on AC and EA = b/3
    • Let T be a point on BA and BT = c/4
    • Let F be a point on BC and FB = a/2
    • Join ET, TF and FE so that make four triangles
    Question
    • Find the area of triangle ETF = ?
    Solution
    Go to Begin

    TR 10 00. Outlines

  • Sine law : a = 2*R*sin(A)
  • Cosine law : a^2 = b^2 + c^2 - 2*b*c*cos(A)
  • ex-circle
      Tangent length AF = s
    • Ex-rdius r = s*tan(A/2)
  • Half angles
    • sin(A/2) = Sqr((s-b)*(s-c)/(b*c)).
    • sin(B/2) = Sqr((s-c)*(s-a)/(c*a)).
    • sin(C/2) = Sqr((s-a)*(s-b)/(a*b)).
  • Half angles
    • cos(A/2) = Sqr(s*(s-a)/(b*c)).
    • cos(B/2) = Sqr(s*(s-b)/(c*a)).
    • cos(C/2) = Sqr(s*(s-c)/(a*b)).
  • Half angles
    • tan(A/2) = Sqr((s-b)*(s-c)/(s*(s-a)).
    • tan(B/2) = Sqr((s-c)*(s-a)/(s*(s-b)).
    • tan(C/2) = Sqr((s-a)*(s-b)/(s*(s-c)).
  • Heron formula : Area of triangle = Sqr(s*(s-a)*(s-b)*(s-c))/(b*c)
  • Incircle
      Tangent length AF = (s - a)
    • in-rdius r = (s - a)*tan(A/2)
  • Sin(A) = 2*Sqr(s*(s-a)*(s-b)*(s-c))/(b*c)
  • Area of triangle
    • Area = b*c*sin(A)/2
    • Area = 2*(R^2)*sin(A)*sin(B)*sin(C)
    • Area = (a*b*c)/(4*R) where R = circum-radius

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