SOLUTION: 2. A parallelogram has diagonals 34 in and 20 in and one side measures 15 in. a. Find the length of the other side b. Find the area c. Find the largest interior angle of the par

                   (a)  Find the length of the other side


The diagonals of a parallelogram bisect each other and divide the parallelogram 
in 4 (four) small triangles.

Let's consider one of four small triangles, formed by two intersecting diagonals and the given side.
This triangle has the sides of 34/2 = 17 in, 20/2 = 10 in and 15 in.

Write the cosine law for this triangle to find the cosine of the angle  'a'  between the diagonals
opposite to the side of 15 inches long

   15^2 = 17^2 + 10^2 - 2*17*10*cos(a),

   cos(a) =  =  = .


The other angle between the diagonals, 'b', is supplementary to angle 'a'. 
Hence, angle 'b' has the cosine  .

Therefore, the square of the side of the parallelogram, opposite to angle 'b', according the cosine law, is

    17^2 + 10^2 - 2*17*10*cos(b) =  = 17^2 + 10^2 + 2*82 = 553.


Hence, the opposite side of the parallelogram to angle 'b' is   = 23.51595203.

     It is the  ANSWER  to question (a).



              (b)  Find the area


For any parallelogram, its diagonals bisect each other
and divide a parallelogram in 4 (four) small triangles.


So, one of such triangles has the sides 34/2 = 17 inches, 20/2 = 10 inches
and 15 inches.


Having three sides of this triangle, we can find its area using the Heron's formula.
For shortness, I will not write the formulas, since they are in each textbook.
I simply will use one of many existing online calculators for it.
So, the area of this specific triangle is  

     = 74.458 in^2.


The entire parallelogram is the union of small triangles.

The interesting fact is that the areas of all these triangles are equal.

You can easy prove it for yourself, if you draw a perpendicular from one of the vertex
of the parallelogram to the opposite diagonal.  This perpendicular then will be the common
altitude of two small triangles. The bases of these small triangle are equal
and the altitude is common - so, the areas of these triangles are the same.
The similar proof works for the opposite vertex and two other triangles.

Thus the area of the parallelogram is 4 times the area of any of small triangles
of the subdivision.  Thus the area of the parallelogram is   = 297.832 in^2.


     It is the  ANSWER  to question (b).



              (c)  Find the largest interior angle of the parallelogram


Let 'C' be the largest interior angle of the parallelogram.

For any two adjacent sides s1 and s2 of a parallelogram, the area of the parallelogram is

    area = s1*s2*sin(C),  where C is the angle concluded between these sides.


We just found the area of the parallelogram in section )b) above and we know that the area is 297.832 in^2.
So, we can write this equation

    297.832 = 15*23.51595203*sin(C).


It gives  sin(C) =  =  0.844340329.


Since 'C' is the largest interior angle, it is obtuse angle, so we can write

    C = 180° - arcsin(0.844340329) = 180° - 57.6013263° = 122.3986737°.


Thus the greatest angle of the parallelogram is about 122.399°.


     It is the  ANSWER  to question (c).

    
Question (c) can be answered/solved in other way.

Write the cosine law for angle C and the opposite to it longest diagonal of 34 inches

    34^2 = 15^2 + 23.51595203^2 - 2*15*23.51595203*cos(C),

    cos(C) =  = -0.535806502.


Hence, 

    C = arccos(-0.535806502)  = 122.399°,

which is consistent with we had above.