# 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.

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