Solve Triangle Given Its Perimeter, Altitute and Angle - Problem With Solution

Solve a triangle, by finding all its sides, given its perimeter, altitude and angle.

Problem :
In the figure below, ABC is a triangle whose perimeter has a length of 100 units and the length of the altitude h is equal to 18 units. The size of the internal angle A is equal to 560 . Find all sides of the triangle.

Solution to Problem :

  • The given perimeter p = 100 gives an equation as follows

    a + b + c = p      (equation 1)

  • The area of the triangle may be calculated using sides c and b as follows

    area = (1 / 2) b c sin (A)

  • But the area of the triangle may also be calculated using the altitude h and corresponding base a as follows

    area = (1 / 2) h a

  • We now combine the two expressions for the area to obtain an equation as follows

    b c sin (A) = h a      (equation 2)

  • A third equation is obtained using the law of cosine as follows

    a 2 = b 2 + c 2 - 2 b c cos (A)      (equation 3)

  • We now have 3 equations with 3 unknowns which we have to solve. Equation (1) gives

    a = p - (b + c)

  • Substitute the above into equation (3) to obtain

    (p - (b + c)) 2 = b 2 + c 2 - 2 b c cos (A)

  • Expand the left hand side of the above equation and simplify

    p 2 + (b + c)) 2 - 2 p (b + c) = b 2 + c 2 - 2 b c cos (A)

    p 2 + b 2 + c 2 + 2 b c - 2 p (b + c) = b 2 + c 2 - 2 b c cos (A)

    p 2 + 2 b c - 2 p (b + c) = - 2 b c cos (A)      (equation 4)

  • We now use a = p - (b + c) in equation (2) and write

    b c sin (A) = h (p - (b + c))

  • which may be written as follows

    b c sin (A) = h p - h (b + c)      (equation 5)

  • We now define two variables as follows.

    Z = b + c and Y = b c

  • and rewrite equations 4 and 5 as follows.

    p 2 + 2 Y - 2 p Z = - 2 Y cos (A)

    Y sin (A) = h p - h Z

  • The above equations make a system of linear equations with unknowns Z and Y.

    - 2 p Z + 2 (1 + cos (A)) Y = - p 2

    h Z + sin (A) Y = h p

  • We now substitute p, h and angle A by their values.

    - 200 Z + Y (2 + 2 cos (56)) = - 10000

    18 Z + Y sin (56) = 1800

  • Solve the above system to obtain

    Z = 62.6456 and Y = 811.035

  • We now substitute Z by b + c and Y by b c to obtain two equations in b and c as follows.

    b + c = 62.6456 and b c = 811.035

  • We now combine the above equations to obtain an equation in one unknown as follows.

    b + 811.035 / b = 62.6456

  • Multiply all terms to obtain a quadratic equation.

    b 2 + 811.035 = 62.6456 b

  • Solve to obtain.

    b = 44.3643 and b = 18.2812

  • We now use the equation b c = 811.035 to find c.

    for b = 44.3643 , c = 18.2812

    and for b = 18.2812, c = 44.3643

  • It is in fact one solution since c and b are interchangeable. let the solution be b = 44.3643 units and c = 18.2812 units and find the third side using equation (3)

    a 2 = b 2 + c 2 - 2 b c cos (A)

    a = sqrt [ 44.3643 2 + 18.2812 2 - 2*44.3643*18.2812 cos (56) ]

    a = 37.3543 units
  • As an example check that the perimeter of the triangle is equal to 100 units.


More references to geometry problems.


Geometry Tutorials, Problems and Interactive Applets.



Popular Pages

More Info