6. given $ \triangle a c p \cong \triangle l n x $, find each missing measure.
a) $ x l = $
b) $ a c = $
c) $ p c = $
d) $ m \angle l = $
e) $ m \angle c = $
f) $ m \angle x = $

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Accepted Answer

# Explanation:
## Step1: Use the property of congruent triangles
Since $\triangle ACP\cong\triangle LNX$, corresponding parts are equal.
- For side lengths:
    - $XL = AP$ (corresponding sides). Given $AP = 13\mathrm{cm}$, so $XL=13\mathrm{cm}$.
    - $AC = LN$ (corresponding sides). Given $LN = 21\mathrm{cm}$, so $AC = 21\mathrm{cm}$.
    - $PC=NX$ (corresponding sides). Since $NX$ is not directly given in a non - side - marked way, but by the congruence of the triangles, we can also use the fact that in congruent triangles, all corresponding elements match.
- For angles:
    - $m\angle L=m\angle A$ (corresponding angles). Given $m\angle A = 29^{\circ}$, so $m\angle L=29^{\circ}$.
    - $m\angle C=m\angle N$. In $\triangle LNX$, since $LN = NX$ (marked as equal), $\triangle LNX$ is isosceles. But since $\triangle ACP\cong\triangle LNX$, and in $\triangle ACP$ we don't have angle - side relations other than through congruence. However, using the angle - sum property of a triangle ($180^{\circ}$ in a triangle) and the fact that $\triangle ACP\cong\triangle LNX$, and if we assume no other markings (except the given side and angle in $\triangle ACP$ and side markings in $\triangle LNX$), we know that $m\angle C=m\angle N$. But since $\triangle LNX$ has $LN = NX$, $m\angle L=m\angle X$ (base angles of isosceles triangle). Wait, no, $\triangle ACP\cong\triangle LNX$, so $m\angle C=m\angle N$, $m\angle A=m\angle L = 29^{\circ}$, $m\angle P=m\angle X$.
    - $m\angle X=m\angle P$. In $\triangle ACP$, using the angle - sum property $m\angle A+m\angle C+m\angle P=180^{\circ}$. In $\triangle LNX$, $m\angle L+m\angle N+m\angle X=180^{\circ}$. Since $m\angle A = m\angle L$ and $m\angle C=m\angle N$, then $m\angle P=m\angle X$. But if we assume that in $\triangle ACP$, we can calculate $m\angle P$ as $180-(29 + m\angle C)$. But since $\triangle ACP\cong\triangle LNX$ and $LN = NX$ (so $m\angle L=m\angle X$ in $\triangle LNX$ if we consider the side - angle relations in $\triangle LNX$ (isosceles triangle with $LN = NX$), but actually, because of congruence $m\angle A=m\angle L = 29^{\circ}$, $m\angle X=m\angle P$. And in $\triangle ACP$, if we assume no other angles are given (except $\angle A$), but using the congruence and the fact that in $\triangle LNX$ (with $LN = NX$) $m\angle L=m\angle X$ (by isosceles triangle property, sides $LN = NX$ implies base angles $\angle L$ and $\angle X$ are equal). So $m\angle X = 29^{\circ}$

# Answer:
a) $XL = 13\mathrm{cm}$
b) $AC=21\mathrm{cm}$
c) $PC = 21\mathrm{cm}$ (since $NX = 21\mathrm{cm}$ as $LN = NX$ and $PC = NX$ by congruence)
d) $m\angle L=29^{\circ}$
e) $m\angle C = 122^{\circ}$ (using angle - sum property in $\triangle ACP$: $m\angle C=180-(29 + 29)=122^{\circ}$)
f) $m\angle X = 29^{\circ}$

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Explanation:

Step1: Use the property of congruent triangles

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