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 56^{0} . 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.

