Question

δbig is similar to δmop. find v, w, x, and y.
figure is not drawn to scale.
v = in.
w =
x =
y = in.

Explanation:

Step1: Use the property of similar triangles for angles

Since \(\triangle BIG\sim\triangle MOP\), corresponding angles are equal.
\(\angle B=\angle M = x^{\circ}\), \(\angle I=\angle O = 117^{\circ}\), \(\angle G=\angle P=22^{\circ}\).
For \(x\):
In \(\triangle BIG\), using the angle - sum property of a triangle (\(\angle B+\angle I+\angle G = 180^{\circ}\)). But since \(\angle B=\angle M\) and we can also note that for similar triangles, \(\angle B=\angle M\).
\(x = 22^{\circ}\) (corresponding angles of similar triangles \(\angle G=\angle M\) is wrong, actually \(\angle B=\angle M\), and in \(\triangle BIG\), \(\angle B=180-(117 + 22)=41^{\circ}\), no, wait, correct way: \(\triangle BIG\sim\triangle MOP\), so \(\angle B=\angle M\), \(\angle I=\angle O = 117^{\circ}\), \(\angle G=\angle P = 22^{\circ}\). Then \(x=\angle B=180-(117 + 22)=41^{\circ}\) is wrong. Wait, no:
Since \(\triangle BIG\sim\triangle MOP\), \(\angle B=\angle M=x\), \(\angle I=\angle O = 117^{\circ}\), \(\angle G=\angle P = 22^{\circ}\). Using the angle - sum property of a triangle (\(x+117 + 22=180\)), \(x = 41^{\circ}\).
\(w=\angle O = 117^{\circ}\) (corresponding angles of similar triangles \(\triangle BIG\sim\triangle MOP\), \(\angle I\) and \(\angle O\) are corresponding).

Step2: Use the property of similar triangles for side - length ratios

The ratio of corresponding sides of similar triangles is equal. \(\frac{BI}{MO}=\frac{IG}{OP}=\frac{BG}{MP}\)
We know that \(\frac{38}{57}=\frac{v}{21}=\frac{30}{y}\)
First, find \(v\):
\(\frac{v}{21}=\frac{38}{57}\), cross - multiply gives \(57v=38\times21\), \(v=\frac{38\times21}{57}\)
\(v = 14\) (since \(38\times21=(2\times19)\times(3\times7)\), \(57 = 3\times19\), \(\frac{38\times21}{57}=\frac{2\times19\times3\times7}{3\times19}=14\))
Then, find \(y\):
\(\frac{30}{y}=\frac{38}{57}\), cross - multiply gives \(38y=30\times57\), \(y=\frac{30\times57}{38}\)
\(y = 45\) (since \(30\times57=(2\times3\times5)\times(3\times19)\), \(38 = 2\times19\), \(\frac{30\times57}{38}=\frac{2\times3\times5\times3\times19}{2\times19}=45\))

Answer:

\(v = 14\) in., \(w = 117^{\circ}\), \(x = 41^{\circ}\), \(y = 45\) in.