Answer :

wegnerkolmp2741o

Answer:

Step-by-step 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

amna04352

Answer:

x = 9

y = 10

Obtuse

Step-by-step explanation:

(x)(x + 16) = (15)(15)

x² + 16x - 225 = 0

x² + 25x - 9x - 225 = 0

x(x + 25) - 9(x + 25) = 0

(x + 25)(x - 9) = 0

x = -25, 9

Since x is a length, can't be negative

x = 9

y(y + 32) = 14(14 + 16)

y² + 32y = 420

y² + 32y - 420 = 0

y² + 42y - 10y - 420 = 0

y(y + 42) - 10(y + 42) = 0

(y + 42)(y - 10) = 0

y = -42, 10

Since y is a length, can't be negative

So y = 10

Using cosine law, find angle ABC

Lengths:

AB = 15

AC = 32 + 10 = 42

BC = 9 + 16 + 14 = 39

42² = 15² + 39² - 2(15)(39)cosB

cosB = 90.88150831 > 90

So, obtuse

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