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B 242.
QUAD is inscribed in circle
P.
If
/_A=x^(2)-15 and
/_Q=100-(1)/(2)x, find
/_Q.

B $242$ . QUAD is inscribed in circle $P$ . If $\angle A=x^{2}-15$ and $\angle Q=100-\frac{1}{2}x$ , find $\angle Q$ .

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Find derivatives of using multiple formulae

Full solution

Q. B

242

. QUAD is inscribed in circle

P

. If

\angle A=x^{2}-15

and

\angle Q=100-\frac{1}{2}x

, find

\angle Q

.

Identify Relationship: Identify the relationship between angles in a circle. $\newline$ In a circle, opposite angles of an inscribed quadrilateral sum up to $180^\circ$ .
Set up Equation: Set up the equation using the relationship. $\newline$ $\angle A + \angle Q = 180$ $\newline$ $(x^2 - 15) + (100 - (\frac{1}{2})x) = 180$
Simplify Equation: Simplify the equation. $\newline$ $x^2 - \frac{1}{2}x + 100 - 15 = 180$ $\newline$ $x^2 - \frac{1}{2}x + 85 = 180$
Solve for x: Solve for x. $\newline$ $x^2 - \frac{1}{2}x - 95 = 0$ $\newline$ Using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $\newline$ Here, $a = 1$ , $b = -\frac{1}{2}$ , $c = -95$ $\newline$ $x = \frac{0.5 \pm \sqrt{(0.5)^2 + 4\cdot1\cdot95}}{2\cdot1}$ $\newline$ $x = \frac{0.5 \pm \sqrt{0.25 + 380}}{2}$ $\newline$ $x = \frac{0.5 \pm \sqrt{380.25}}{2}$ $\newline$ $x = \frac{0.5 \pm 19.5}{2}$ $\newline$ $x = 10$ or $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $0$
Substitute x: Substitute $x$ back into the expression for $/_Q$ .
$/_Q = 100 - (1/2)x$
For $x = 10$ , $/_Q = 100 - (1/2)\cdot10 = 100 - 5 = 95$
For $x = -19$ , $/_Q = 100 - (1/2)\cdot(-19) = 100 + 9.5 = 109.5$

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