Question

15 in $\triangle abc$ below, $\overline{ac}$ is extended through $c$ to $d$, $m\angle a = (3x - 22)\degree$, $m\angle b = (4x - 18)\degree$, and $m\angle bcd = (6x - 23)\degree$.
determine and state $m\angle acb$.
16 in the diagram below of triangle $abc$, $\overline{ac}$ is extended through point $c$ to point $d$, and $be$ is drawn to $ac$.
which equation is always true?

1. $m\angle 1 = m\angle 3 + m\angle 2$
2. $m\angle 5 = m\angle 3 - m\angle 2$
3. $m\angle 6 = m\angle 3 - m\angle 2$
4. $m\angle 7 = m\angle 3 + m\angle 2$

17 in the diagram below of $\triangle bcd$, side $\overline{db}$ is extended to point $a$.
which statement must be true?

1. $m\angle c > m\angle d$
2. $m\angle abc < m\angle d$
3. $m\angle abc > m\angle c$
4. $m\angle abc > m\angle c + m\angle d$

18 side $\overline{pq}$ of $\triangle pqr$ is extended through $q$ to $t$. which statement is not always true?

1. $m\angle rqt > m\angle r$
2. $m\angle rqt > m\angle p$
3. $m\angle rqt = m\angle p + m\angle r$
4. $m\angle rqt > m\angle pqr$

19 in $\triangle abc$, an exterior angle at $c$ measures $50$, $m\angle a > 30$. which inequality must be true?

1. $m\angle b < 20$
2. $m\angle b > 20$
3. $m\angle bca < 130$
4. $m\angle bca > 130$

20 in all isosceles triangles, the exterior angle base angle must always be

1. a right angle
2. an acute angle
3. an obtuse angle
4. equal to the vertex angle

21 if one exterior angle of a triangle is acute triangle must be

1. right
2. acute
3. obtuse
4. equiangular

Explanation:

Problem 15

Step1: Use exterior angle theorem

The exterior angle of a triangle equals the sum of the two remote interior angles, so:
$$(6x - 23) = (3x - 22) + (4x - 18)$$

Step2: Simplify and solve for $x$

$$6x - 23 = 7x - 40$$
$$-23 + 40 = 7x - 6x$$
$$x = 17$$

Step3: Find $\text{m}\angle BCD$

Substitute $x=17$ into the exterior angle expression:
$$\text{m}\angle BCD = 6(17) - 23 = 102 - 23 = 79^\circ$$

Step4: Calculate $\text{m}\angle ACB$

$\angle ACB$ and $\angle BCD$ are supplementary (linear pair):
$$\text{m}\angle ACB = 180^\circ - 79^\circ = 101^\circ$$

Use the exterior angle theorem for triangles. $\angle 1$ is an exterior angle of $\triangle ABC$, so $\text{m}\angle 1 = \text{m}\angle 3 + \text{m}\angle 2$ is false. $\angle 5$ cannot be expressed as $\text{m}\angle 3 - \text{m}\angle 2$. $\angle 6$ is an exterior angle of $\triangle BEC$, so $\text{m}\angle 6 = \text{m}\angle 3 + \text{m}\angle 2$ is false. $\angle 7$ is an exterior angle of $\triangle BEC$, so $\text{m}\angle 7 = \text{m}\angle 3 + \text{m}\angle 2$ is always true.

$\angle ABC$ is an exterior angle of $\triangle BCD$, so by the exterior angle theorem, $\text{m}\angle ABC = \text{m}\angle C + \text{m}\angle D$. This means $\text{m}\angle ABC > \text{m}\angle D$, $\text{m}\angle ABC > \text{m}\angle C$, and $\text{m}\angle ABC$ cannot be greater than $\text{m}\angle C + \text{m}\angle D$. Only $\text{m}\angle C > \text{m}\angle D$ is not guaranteed, but the theorem confirms $\text{m}\angle ABC > \text{m}\angle D$ is false, and $\text{m}\angle ABC > \text{m}\angle C$ is true.

Answer:

$\boldsymbol{101^\circ}$

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Problem 16