Question

10 in the diagram below of $\triangle abc$, $\overline{bc}$ is extended to $d$.
(not drawn to scale)
if $m\angle a = x^2 - 6x$, $m\angle b = 2x - 3$, and $m\angle acd = 9x + 27$, what is the value of $x$

1. 10
2. 2
3. 3
4. 15

11 in the diagram of $\triangle pqr$ shown below, $\overline{pr}$ is extended to $s$, $m\angle p = 110$, $m\angle q = 4x$, and $m\angle qrs = x^2 + 5x$.
what is $m\angle q$

1. 44
2. 40
3. 11
4. 10

12 in $\triangle abc$, $m\angle cab = 2x$ and $m\angle acb = x + 30$. if $\overline{ab}$ is extended through point $b$ to point $d$, $m\angle cbd = 5x - 50$. what is the value of $x$

1. 25
2. 30
3. 40
4. 46

13 angle $kml$ is the vertex angle of isosceles triangle $klm$ below. side $lm$ is extended through vertex $m$ to point $n$.
if $m\angle k = 15^\circ$, determine and state $m\angle kmn$
14 in the diagram below of $\triangle hqp$, side $\overline{hp}$ is extended through $p$ to $t$, $m\angle qpt = 6x + 20$, $m\angle hqp = x + 40$, and $m\angle phq = 4x - 5$. find $m\angle qpt$.
(not drawn to scale)

Explanation:

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Problem 10

Step1: Apply exterior angle theorem

The exterior angle of a triangle equals the sum of the two remote interior angles:
$$m\angle ACD = m\angle A + m\angle B$$
Substitute the given expressions:
$$9x + 27 = (x^2 - 6x) + (2x - 3)$$

Step2: Simplify and solve quadratic

Combine like terms and rearrange into standard quadratic form:
$$x^2 - 13x - 30 = 0$$
Factor the quadratic:
$$(x - 15)(x + 2) = 0$$
Solve for $x$: $x=15$ or $x=-2$. Since angle measures cannot be negative, $x=15$.
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Problem 11

Step1: Apply exterior angle theorem

$$m\angle QRS = m\angle P + m\angle Q$$
Substitute the given expressions:
$$x^2 + 5x = 110 + 4x$$

Step2: Simplify and solve quadratic

Rearrange into standard form:
$$x^2 + x - 110 = 0$$
Factor the quadratic:
$$(x + 11)(x - 10) = 0$$
Solve for $x$: $x=10$ or $x=-11$. Use positive $x=10$.

Step3: Calculate $m\angle Q$

$$m\angle Q = 4x = 4(10) = 40$$
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Problem 12

Step1: Apply exterior angle theorem

$$m\angle CBD = m\angle CAB + m\angle ACB$$
Substitute the given expressions:
$$5x - 50 = 2x + (x + 30)$$

Step2: Simplify and solve for $x$

Combine like terms:
$$5x - 50 = 3x + 30$$
$$2x = 80$$
$$x = 40$$
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Problem 13

Step1: Identify base angles of isosceles triangle

Since $\angle KML$ is the vertex angle, $\angle K = \angle L = 15^\circ$.

Step2: Calculate $\angle KML$

Sum of angles in a triangle is $180^\circ$:
$$m\angle KML = 180^\circ - 15^\circ - 15^\circ = 150^\circ$$

Step3: Find $\angle KMN$

$\angle KMN$ and $\angle KML$ are supplementary (linear pair):
$$m\angle KMN = 180^\circ - 150^\circ = 30^\circ$$
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Problem 14

Step1: Apply exterior angle theorem

$$m\angle QPT = m\angle HOP + m\angle PHQ$$
Substitute the given expressions:
$$6x + 20 = (x + 40) + (4x - 5)$$

Step2: Simplify and solve for $x$

Combine like terms:
$$6x + 20 = 5x + 35$$
$$x = 15$$

Step3: Calculate $m\angle QPT$

$$m\angle QPT = 6(15) + 20 = 110$$
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Answer:

1. Problem 10: 4) 15
2. Problem 11: 2) 40
3. Problem 12: 3) 40
4. Problem 13: $30^\circ$
5. Problem 14: $110^\circ$