Suppose that we have a right triangle $abc$ with the right angle at $b$ such that $ac = \sqrt{61}$ and $ab = 5.$ a circle is drawn with its center on $ab$ such that the circle is tangent to $ac$ and $bc.$ if $p$ is the point where the circle and side $ac$ meet, then what is $cp$?

  CP = 6

Step-by-step explanation:

The length of segment BC is given by the Pythagorean theorem:

  AC² = AB² +BC²

  (√61)² = 5² + BC² . . . . . fill in the given numbers

  61 -25 = BC² = 36 . . . . .subtract 25

  BC = 6 . . . . . . . . . . . . . . take the square root

Since the center of the circle is on AB and is tangent to BC, it must pass through point B. That is, segment BC of length 6 is one of the tangent lines from point C. The other one, to point P, must be the same length, so ...

  CP = 6

(x+3)(x+7)

step-by-step explanation:

(2x + 16) + (x) = 180

step-by-step explanation:

given,

abcd is a quadrilateral inscribed in a circle

angle a is labeled x plus 16, (x + 16)

angle b is labeled x degrees, (x)

angle c is labeled 6x minus 4, (6x – 4)

angle d is labeled 2x plus 16, (2x + 16)

what equation can be used to solve for angle b?

we know that the sum of the opposite angles of a simple quadrilateral inscribed in a circle is 180 degrees.

angle b is opposite angle d

therefore, angle b + angle d = 180

(x) + (2x + 16) = 180 

same as (2x + 16) + (x) = 180