Answer:
-2 < x < 32.3
Explanation:
You want to know the range of the possible values of x in the given figure with two adjacent triangles.
Solution
Angle ABC will be the largest it can be when ∆ABC is isosceles. That is, AB=CB=DB. The law of cosines can be used to find the angle.
Let AB=CB=DB=a, then the law of cosines tells us for ∆CBD ...
14² = a² +a² -2a·a·cos(37°) = 2a²(1 -cos(37°))
Similarly, for ∆ABC, we have ...
13² = 2a²(1 -cos(x+2)°)
The ratio of these two equations is ...
13²/14² = (2a²(1 -cos(x+2)°))/(2a²(1 -cos(37°))
1 -cos(x +2)° = (13/14)²(1 -cos(37°)) . . . . . multiply by (1-cos(37°))
cos(x +2)° = 1 -(13/14)²(1 -cos(37°)) . . . . . solve for cos(x+2)°
x = arccos(1 -(13/14)²(1 -cos(37°))) -2° . . . . solve for x
x ≈ 32.3° . . . . . maximum value of x
The minimum value of x will be -2, when angle ABC is 0°. This can happen at extreme lengths of BC (long or short).
-2 < x < 32.3