Question

1. what is the length of the missing side, x? 40 in. 51° 35 in. 74° 74° 49 in. 55° x

Explanation:

Step1: Verify triangle similarity

First, check if the two triangles are similar. For the first triangle, the angles are \(51^\circ\), \(74^\circ\), and \(180 - 51 - 74 = 55^\circ\). The second triangle has angles \(74^\circ\), \(55^\circ\), and \(180 - 74 - 55 = 51^\circ\). So, the triangles are similar by AA (Angle - Angle) similarity criterion.

Step2: Identify corresponding sides

In similar triangles, corresponding sides are proportional. Let's identify the corresponding sides. The side of length \(40\) in the first triangle corresponds to the side of length \(49\) in the second triangle? Wait, no. Wait, the first triangle has sides: let's see, angle \(51^\circ\) is opposite \(35\) in, angle \(74^\circ\) is opposite \(40\) in, and angle \(55^\circ\) is opposite... Wait, no. Wait, let's list the angles:

First triangle angles: \(51^\circ\), \(74^\circ\), \(55^\circ\)

Second triangle angles: \(51^\circ\), \(74^\circ\), \(55^\circ\) (since \(180 - 74 - 55 = 51\))

So, the side opposite \(51^\circ\) in first triangle: let's see, in first triangle, sides: \(40\) in (opposite \(74^\circ\)? Wait, no. Wait, first triangle: angle \(51^\circ\), side adjacent? Wait, maybe better to match angles.

Angle \(74^\circ\) in first triangle: side opposite? Wait, first triangle: sides are \(40\) in, \(35\) in, and the third side. Wait, maybe I made a mistake. Wait, first triangle: angles \(51^\circ\), \(74^\circ\), \(55^\circ\). So, the side opposite \(51^\circ\) is \(35\) in? Wait, no. Wait, first triangle: let's label the angles. Let's say in first triangle, angle \(A = 51^\circ\), angle \(B = 74^\circ\), angle \(C = 55^\circ\). Then side \(a\) (opposite angle \(A\)) is \(35\) in, side \(b\) (opposite angle \(B\)) is \(40\) in, side \(c\) (opposite angle \(C\)) is? Wait, no, the first triangle has sides: one side is \(40\) in, one is \(35\) in, and the included angles? Wait, maybe the first triangle: angle \(51^\circ\) is between sides \(40\) and the side opposite \(35\)? Wait, maybe I should use the Law of Sines.

Law of Sines: \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\) for a triangle with sides \(a,b,c\) opposite angles \(A,B,C\) respectively.

For the first triangle:

Let’s denote:

Angle \(A_1 = 51^\circ\), angle \(B_1 = 74^\circ\), angle \(C_1 = 55^\circ\)

Sides: \(a_1\) (opposite \(A_1\)) =?, \(b_1\) (opposite \(B_1\)) = \(40\) in, \(c_1\) (opposite \(C_1\)) = \(35\) in? Wait, no, the first triangle has sides: \(40\) in, \(35\) in, and the third side. Wait, maybe the first triangle: side of length \(40\) is opposite angle \(55^\circ\)? Wait, no, let's calculate using Law of Sines.

In first triangle:

\(\frac{40}{\sin 55^\circ}=\frac{35}{\sin 51^\circ}\)? Wait, no, let's check:

Wait, first triangle angles: \(51^\circ\), \(74^\circ\), \(55^\circ\). Let's assign:

Let angle \(A = 51^\circ\), angle \(B = 74^\circ\), angle \(C = 55^\circ\)

Then side \(a\) (opposite \(A\)): let's say side \(a = 35\) in (since it's adjacent to \(51^\circ\) and \(74^\circ\)? Wait, maybe the first triangle has sides: \(40\) in (opposite \(74^\circ\)), \(35\) in (opposite \(51^\circ\)), and the third side (opposite \(55^\circ\)): let's calculate using Law of Sines.

\(\frac{40}{\sin 74^\circ}=\frac{35}{\sin 51^\circ}\)

Calculate \(\sin 74^\circ\approx 0.9613\), \(\sin 51^\circ\approx 0.7771\)

Left side: \(\frac{40}{0.9613}\approx 41.61\)

Right side: \(\frac{35}{0.7771}\approx 45.04\). Wait, that's not equal. Wait, maybe I assigned the sides wrong.

Wait, maybe the first triangle: the side of length \(40\) is opposite \(55^\circ\), and the side of length \(35\) is opposite \(74^\circ\)? Let's check:

\(\frac{40}{\sin 55^\circ}=\frac{35}{\sin 74^\circ}\)

\(\sin 55^\circ\approx 0.8192\), \(\sin 74^\circ\approx 0.9613\)

Left side: \(\frac{40}{0.8192}\approx 48.83\)

Right side: \(\frac{35}{0.9613}\approx 36.41\). Not equal.

Wait, maybe the first triangle: angles are \(51^\circ\), \(74^\circ\), \(55^\circ\), and the sides: \(40\) in (between \(51^\circ\) and \(74^\circ\)), \(35\) in (between \(74^\circ\) and \(55^\circ\)), and the third side (between \(55^\circ\) and \(51^\circ\)). Then, using Law of Sines:

Let’s denote:

\(a = 40\) (between \(51^\circ\) and \(74^\circ\), so opposite \(55^\circ\))

\(b = 35\) (between \(74^\circ\) and \(55^\circ\), so opposite \(51^\circ\))

\(c\) (between \(55^\circ\) and \(51^\circ\), opposite \(74^\circ\))

Then by Law of Sines: \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\)

So \(\frac{40}{\sin 55^\circ}=\frac{35}{\sin 51^\circ}=\frac{c}{\sin 74^\circ}\)

Calculate \(\frac{40}{\sin 55^\circ}\approx \frac{40}{0.8192}\approx 48.83\)

\(\frac{35}{\sin 51^\circ}\approx \frac{35}{0.7771}\approx 45.04\). Wait, that's a problem. Maybe the triangles are not similar? Wait, no, the angles are the same, so they must be similar. So maybe my side assignment is wrong.

Wait, the second triangle: angles \(74^\circ\), \(55^\circ\), \(51^\circ\), sides: \(49\) in (let's see, which angle is opposite \(49\) in? Let's say angle \(51^\circ\) is opposite \(49\) in? Wait, no. Wait, the second triangle has side \(x\) opposite \(74^\circ\), side \(49\) in opposite \(51^\circ\), and the other side opposite \(55^\circ\).

Wait, maybe the correct correspondence is:

First triangle: angle \(51^\circ\), side opposite is \(35\) in; angle \(74^\circ\), side opposite is \(40\) in; angle \(55^\circ\), side opposite is?

Second triangle: angle \(51^\circ\), side opposite is \(49\) in; angle \(74^\circ\), side opposite is \(x\); angle \(55^\circ\), side opposite is?

Wait, no, that can't be. Wait, let's recalculate the angles.

First triangle: angles \(51^\circ\), \(74^\circ\), so third angle is \(180 - 51 - 74 = 55^\circ\). Correct.

Second triangle: angles \(74^\circ\), \(55^\circ\), so third angle is \(180 - 74 - 55 = 51^\circ\). Correct. So angles are same, so triangles are similar. Therefore, the ratio of corresponding sides should be equal.

So, let's list the angles and their opposite sides:

First triangle:

- Angle \(51^\circ\): opposite side \(35\) in

- Angle \(74^\circ\): opposite side \(40\) in

- Angle \(55^\circ\): opposite side (let's call it \(s\))

Second triangle:

- Angle \(51^\circ\): opposite side \(49\) in

- Angle \(74^\circ\): opposite side \(x\)

- Angle \(55^\circ\): opposite side (let's call it \(y\))

Since the triangles are similar, the ratio of sides opposite equal angles should be equal.

So, \(\frac{\text{Side opposite }51^\circ\text{ in first triangle}}{\text{Side opposite }51^\circ\text{ in second triangle}}=\frac{\text{Side opposite }74^\circ\text{ in first triangle}}{\text{Side opposite }74^\circ\text{ in second triangle}}\)

So, \(\frac{35}{49}=\frac{40}{x}\)

Wait, is that the correct correspondence? Wait, no. Wait, maybe the side of length \(35\) in first triangle corresponds to the side of length \(49\) in second triangle? No, that would be if angle \(51^\circ\) is between them. Wait, maybe I mixed up the sides.

Wait, let's do it properly. Let's denote the first triangle as \(\triangle ABC\) with \(A = 51^\circ\), \(B = 74^\circ\), \(C = 55^\circ\), and sides \(a = BC = 35\) in, \(b = AC = 40\) in, \(c = AB\) (unknown).

Second triangle as \(\triangle DEF\) with \(D = 51^\circ\), \(E = 74^\circ\), \(F = 55^\circ\), and sides \(d = EF = x\), \(e = DF = 49\) in, \(f = DE\) (unknown).

By AA similarity, \(\triangle ABC \sim \triangle DEF\) (order of angles: \(A = D\), \(B = E\), \(C = F\))

Therefore, the ratio of corresponding sides: \(\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}\)

So, \(AB\) corresponds to \(DE\), \(BC\) corresponds to \(EF\), \(AC\) corresponds to \(DF\)

So, \(BC = 35\) in (side \(a\)) corresponds to \(EF = x\) (side \(d\))? Wait, no, \(BC\) is opposite \(A\) (\(51^\circ\)), \(EF\) is opposite \(D\) (\(51^\circ\)), so \(BC\) corresponds to \(EF\). Wait, \(BC\) is side \(a\) (opposite \(A\)), \(EF\) is side \(d\) (opposite \(D\)), so \(\frac{BC}{EF}=\frac{AC}{DF}\)

So, \(BC = 35\), \(EF = x\), \(AC = 40\), \(DF = 49\)

Therefore, \(\frac{35}{x}=\frac{40}{49}\)

Wait, no, that would be \(\frac{BC}{EF}=\frac{AC}{DF}\) => \(\frac{35}{x}=\frac{40}{49}\) => \(x=\frac{35\times 49}{40}\). Wait, that gives \(x = \frac{1715}{40}=42.875\). But that doesn't seem right. Wait, maybe the correspondence is different.

Wait, maybe \(AC\) (40 in) corresponds to \(EF\) (x), and \(BC\) (35 in) corresponds to \(DF\) (49 in). Let's check:

\(\frac{BC}{DF}=\frac{AC}{EF}\) => \(\frac{35}{49}=\frac{40}{x}\)

Cross - multiply: \(35x = 40\times49\)

\(35x = 1960\)

\(x=\frac{1960}{35}=56\)

Ah, that makes sense. Let's verify:

\(\frac{35}{49}=\frac{5}{7}\), \(\frac{40}{56}=\frac{5}{7}\). Yes, that's correct. So the ratio is \(\frac{5}{7}\).

So, the mistake was in the correspondence of the sides. The side of length \(35\) in the first triangle corresponds to the side of length \(49\) in the second triangle, and the side of length \(40\) in the first triangle corresponds to the side of length \(x\) in the second triangle. So, the proportion is \(\frac{35}{49}=\frac{40}{x}\)

Step3: Solve for \(x\)

From \(\frac{35}{49}=\frac{40}{x}\), cross - multiply:

\(35x = 40\times49\)

\(35x = 1960\)

Divide both sides by \(35\):

\(x=\frac{1960}{35}=56\)

Answer:

The length of the missing side \(x\) is \(\boldsymbol{56}\) inches.