Question

find the perimeter of $\triangle vwx$. round your answer to the nearest tenth if necessary. figures are not necessarily drawn to scale.
triangle $stu$ with sides $st=15$, $tu=19$, $us=20$ and angles $\angle s=63^\circ$, $\angle t=72^\circ$, $\angle u=45^\circ$; triangle $vwx$ with angles $\angle v=63^\circ$, $\angle w=72^\circ$, $\angle x=45^\circ$, side $vw=42$, side $vx=56$, side $wx=x$

Explanation:

Step1: Identify Similar Triangles

Triangles \( \triangle STU \) and \( \triangle VWX \) have the same angle measures (\( 63^\circ \), \( 72^\circ \), \( 45^\circ \)), so they are similar by AA (Angle - Angle) similarity criterion.

Step2: Find the Scale Factor

In \( \triangle STU \), sides are \( 15 \), \( 19 \), \( 20 \). In \( \triangle VWX \), one side corresponding to \( 15 \) in \( \triangle STU \) is \( 42 \)? Wait, no. Wait, let's match the angles. Angle at \( S \) is \( 63^\circ \), angle at \( T \) is \( 72^\circ \), angle at \( U \) is \( 45^\circ \). In \( \triangle VWX \), angle at \( V \) is \( 63^\circ \), angle at \( W \) is \( 72^\circ \), angle at \( X \) is \( 45^\circ \). So side \( ST = 15 \) (opposite \( 45^\circ \) in \( \triangle STU \)), side \( VW = 42 \) (opposite \( 45^\circ \) in \( \triangle VWX \))? Wait, no, let's check the sides. Wait, in \( \triangle STU \), side \( ST = 15 \), \( TU = 19 \), \( US = 20 \). In \( \triangle VWX \), side \( VW = 42 \), \( WX=x \), \( XV = 56 \). Let's find the ratio of corresponding sides. Let's see, angle \( S = 63^\circ \), angle \( V = 63^\circ \); angle \( T = 72^\circ \), angle \( W = 72^\circ \); angle \( U = 45^\circ \), angle \( X = 45^\circ \). So \( \triangle STU \sim \triangle VWX \) with correspondence \( S ightarrow V \), \( T ightarrow W \), \( U ightarrow X \). So side \( ST \) (length \( 15 \)) corresponds to side \( VW \) (length \( 42 \))? Wait, no, \( ST \) is between \( S(63^\circ) \) and \( T(72^\circ) \), \( VW \) is between \( V(63^\circ) \) and \( W(72^\circ) \), so \( ST \) corresponds to \( VW \), \( TU \) corresponds to \( WX \), \( US \) corresponds to \( XV \). So the ratio of \( VW \) to \( ST \) is \( \frac{42}{15}=\frac{14}{5} = 2.8 \). Let's check with \( XV = 56 \) and \( US = 20 \). \( \frac{56}{20}=\frac{14}{5}=2.8 \), so the scale factor \( k = 2.8 \).

Step3: Find the Length of \( WX \)

In \( \triangle STU \), \( TU = 19 \). Since \( TU \) corresponds to \( WX \) in \( \triangle VWX \), we have \( WX=TU\times k \). So \( WX = 19\times2.8=53.2 \).

Step4: Calculate the Perimeter of \( \triangle VWX \)

The sides of \( \triangle VWX \) are \( VW = 42 \), \( WX = 53.2 \), \( XV = 56 \). The perimeter \( P=42 + 53.2+56 \).

\( P=42+53.2 + 56=151.2 \)

Wait, let's re - check the correspondence. Maybe I mixed up the sides. Let's use the Law of Sines. For \( \triangle STU \), by Law of Sines: \( \frac{ST}{\sin U}=\frac{TU}{\sin S}=\frac{US}{\sin T} \)

\( \frac{15}{\sin45^\circ}=\frac{19}{\sin63^\circ}=\frac{20}{\sin72^\circ} \)

For \( \triangle VWX \), by Law of Sines: \( \frac{VW}{\sin X}=\frac{WX}{\sin V}=\frac{XV}{\sin W} \)

\( \sin X=\sin45^\circ \), \( \sin V=\sin63^\circ \), \( \sin W=\sin72^\circ \)

We know \( VW = 42 \), \( XV = 56 \)

From \( \triangle STU \): \( \frac{ST}{\sin45^\circ}=\frac{15}{\sin45^\circ}\approx\frac{15}{0.7071}\approx21.21 \)

\( \frac{TU}{\sin63^\circ}=\frac{19}{\sin63^\circ}\approx\frac{19}{0.8910}\approx21.32 \) (close, due to rounding)

\( \frac{US}{\sin72^\circ}=\frac{20}{\sin72^\circ}\approx\frac{20}{0.9511}\approx21.03 \) (close, since it's a triangle, the Law of Sines ratios should be equal, slight differences due to rounding)

For \( \triangle VWX \): \( \frac{VW}{\sin X}=\frac{42}{\sin45^\circ}\approx\frac{42}{0.7071}\approx59.39 \)

\( \frac{XV}{\sin W}=\frac{56}{\sin72^\circ}\approx\frac{56}{0.9511}\approx58.88 \) (close, due to rounding)

Let's use the ratio of corresponding sides. Let's take \( US = 20 \) and \( XV = 56 \). The ratio \( \frac{XV}{US}=\frac{56}{20}=2.8 \)

\( ST = 15 \), so corresponding side \( VW \) should be \( 15\times2.8 = 42 \) (which matches). \( TU = 19 \), so corresponding side \( WX=19\times2.8 = 53.2 \)

Now, perimeter of \( \triangle VWX \) is \( 42 + 56+53.2=151.2 \)

Answer:

\( 151.2 \)