Answer:
[tex]1<x<4[/tex]
Step-by-step explanation:
Part 1) Find the value of x
we know that
The triangle ABC is an isosceles triangle, because has two equal sides
AB=BC
so
∠A=∠C
Remember that the sum of the interior angles in any triangle must ne equal to 180 degrees
so
[tex]A+B+C=180^o[/tex]
we have
[tex]B=35^o+20^o=55^o[/tex]
Find the measure of angle A
[tex]2A+55^o=180^o[/tex]
[tex]2A=125^o[/tex]
[tex]A=62.5^o[/tex]
therefore
[tex]B=62.5^o[/tex]
Applying the law of sines In the triangle ABD
[tex]\frac{12}{sin(35^o)}=\frac{BD}{sin(62.5^o)}[/tex]
solve for BD
[tex]BD=\frac{12}{sin(35^o)}sin(62.5^o)[/tex] ----> equation A
Applying the law of sines In the triangle CDB
[tex]\frac{4x-4}{sin(20^o)}=\frac{BD}{sin(62.5^o)}[/tex]
solve for BD
[tex]BD=\frac{4x-4}{sin(20^o)}sin(62.5^o)[/tex] ----> equation B
equate equation A and equation B
[tex]\frac{12}{sin(35^o)}sin(62.5^o)=\frac{4x-4}{sin(20^o)}sin(62.5^o)[/tex]
solve for x
[tex]12sin(20^o)=(4x-4)sin(35^o)[/tex]
[tex]12sin(20^o)=(4x)sin(35^o)-4sin(35^o)[/tex]
[tex](4x)sin(35^o)=12sin(20^o)+4sin(35^o)[/tex]
[tex]x=[12sin(20^o)+4sin(35^o)]/(4sin(35^o))[/tex]
[tex]x=2.8\ units[/tex]
Part 2) Find the range of values of x
Remember that
(4x-4) must be greater than zero
so
[tex]4x-4>0[/tex]
[tex]4x>4\\x>1[/tex]
we know that
In any triangle: The shortest side is always opposite the smallest interior angle. The longest side is always opposite the largest interior angle
That means
[tex]4x-4< 12[/tex]
solve for x
[tex]4x< 16\\x<4[/tex]
therefore
[tex]1<x<4[/tex]
