Let x=ab=ac, and y=bc, and z=ad.
Since the perimeter of the triangle abc is 36, you have:
Perimeter of abc = 36
ab + ac + bc = 36
x + x + y = 36
(eq. 1) 2x + y = 36
The triangle is isosceles (it has two sides with equal length: ab and ac). The line perpendicular to the third side (bc) from the opposite vertex (a), divides that third side into two equal halves: the point d is the middle point of bc. This is a property of isosceles triangles, which is easily shown by similarity.
Hence, we have that bd = dc = bc/2 = y/2 (remember we called bc = y).
The perimeter of the triangle abd is 30:
Permiter of abd = 30
ab + bd + ad = 30
x + y/2 + z =30
(eq. 2) 2x + y + 2z = 60
So, we have two equations on x, y and z:
(eq.1) 2x + y = 36
(eq.2) 2x + y + 2z = 60
Substitute 2x + y by 36 from (eq.1) in (eq.2):
(eq.2') 36 + 2z = 60
And solve for z:
36 + 2z = 60 => 2z = 60 - 36 => 2z = 24 => z = 12
The measure of ad is 12.
If you prefer a less algebraic reasoning:
- The perimeter of abd is half the perimeter of abc plus the length of ad (since you have "cut" the triangle abc in two halves to obtain the triangle abd).
- Then, ad is the difference between the perimeter of abd and half the perimeter of abc:
ad = 30 - (36/2) = 30 - 18 = 12
40 is the answer if your asking for the second angle
Answer:
Step-by-step explanation:
So you should know. Can you give us a better picture?
Answer:
The solutions are linearly independent because the Wronskian is not equal to 0 for all x.
The value of the Wronskian is 
Step-by-step explanation:
We can calculate the Wronskian using the fundamental solutions that we are provided and their corresponding the derivatives, since the Wroskian is defined as the following determinant.
Thus replacing the functions of the exercise we get:
Working with the determinant we get
Thus we have found that the Wronskian is not 0, so the solutions are linearly independent.
