Answer:
Explanation:
First we will find the length of y
Two secants intersect outside the circle.
The outside piece times the total length on one side equals the outside length times the total length on the other side
14* 30 = y* (32+y)
420 =32y + y^2
Subtract 420
0 =y^2 +32y -420
Solving for y
Factor
0 = (y+42)(y-10)
y = -42 y= 10
Y can't be negative so y=10
Now we find x
A secant and a tangent intersect outside the circle.
BA ^2 = Outside length * Total length
15^2 = x* (x+16)
225 = x^2 +16x
Subtract 225
0 = x^2 +16x-225
Solving for x
0=(x-9) (x+25)
x=9 x=-25
Since we can't be negative x=9
Now we can use the Pythagorean theorem to see if the triangle is right acute or obtuse
(15)^2 + (9+16+14) ^2 vs42^2 (32+10)^2
225 + 39^2 vs 42^2
225+1521 1764
1746 <1764
Obtuse triangle