Question

find ( x ). round all values to the nearest tenth to calculate the next value. round your final answer to the tenth as well.

Explanation:

Step1: Find the height (h) of the left triangle

In the left right - triangle with hypotenuse 5 and angle \(57^{\circ}\), we use the sine function. The sine of an angle in a right - triangle is defined as \(\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}\). Let the height be \(h\). So \(\sin(57^{\circ})=\frac{h}{5}\), then \(h = 5\times\sin(57^{\circ})\).
We know that \(\sin(57^{\circ})\approx0.8387\), so \(h = 5\times0.8387 = 4.1935\approx4.2\) (rounded to the nearest tenth).

Step2: Find \(x\) using the right - triangle on the right

In the right - triangle on the right, we have a \(45^{\circ}-45^{\circ}-90^{\circ}\) triangle? Wait, no, it's a right - triangle with angle \(45^{\circ}\) and the height \(h\) (which is also the opposite side to the non - right angle in the left triangle and the adjacent side to the \(45^{\circ}\) angle in the right triangle? Wait, no, in the right - triangle on the right, the height \(h\) is the opposite side to the \(45^{\circ}\) angle? Wait, no, let's re - examine. The right - triangle on the right has angle \(45^{\circ}\), right angle, and the side we want to find is \(x\) (the hypotenuse? No, wait, the side adjacent to the \(45^{\circ}\) angle? Wait, no, in the right - triangle on the right, the height \(h\) is equal to the other leg (because \(\tan(45^{\circ}) = 1=\frac{h}{x}\)? Wait, no, if the angle is \(45^{\circ}\) and the triangle is right - angled, then it is an isosceles right - triangle, so the two legs are equal. Wait, the height \(h\) is one leg, and \(x\) is the other leg? Wait, no, let's use the tangent function. Wait, no, in the right - triangle on the right, the angle is \(45^{\circ}\), the height \(h\) is the opposite side, and \(x\) is the adjacent side? Wait, no, \(\tan(45^{\circ})=\frac{h}{x}\), but \(\tan(45^{\circ}) = 1\), so \(h=x\)? Wait, that can't be. Wait, no, I made a mistake. Let's re - do step 1.

Wait, in the left right - triangle, the angle is \(57^{\circ}\), hypotenuse is 5, and the height \(h\) is the opposite side to the \(57^{\circ}\) angle. So \(\sin(57^{\circ})=\frac{h}{5}\), so \(h = 5\sin(57^{\circ})\approx5\times0.8387 = 4.1935\approx4.2\).

Now, in the right - triangle on the right, we have a right - triangle with angle \(45^{\circ}\), and the height \(h\) is the opposite side to the \(45^{\circ}\) angle? No, wait, the right - triangle on the right has angle \(45^{\circ}\), and the side \(x\) is the hypotenuse? No, wait, the right - triangle on the right: the height \(h\) is one leg, and \(x\) is the other leg. Since \(\tan(45^{\circ})=\frac{h}{x}\), and \(\tan(45^{\circ}) = 1\), so \(x = h\)? But that's not correct. Wait, no, maybe the height \(h\) is the adjacent side to the \(45^{\circ}\) angle. Wait, no, let's use the sine function. In the right - triangle on the right, \(\sin(45^{\circ})=\frac{h}{x}\), so \(x=\frac{h}{\sin(45^{\circ})}\).

We know that \(\sin(45^{\circ})=\frac{\sqrt{2}}{2}\approx0.7071\). We found \(h\approx4.2\). Then \(x=\frac{4.2}{0.7071}\approx5.94\approx5.9\)? Wait, no, wait, let's recalculate \(h\) more accurately.

\(\sin(57^{\circ})\approx0.8386705679454239\), so \(h = 5\times0.8386705679454239=4.1933528397271195\approx4.2\) (rounded to the nearest tenth).

Now, in the right - triangle with angle \(45^{\circ}\), the two legs are equal (because \(\tan(45^{\circ}) = 1\)), so if the height \(h\) is one leg, then the other leg (which is \(x\)) is equal to \(h\) only if it's a \(45 - 45-90\) triangle. Wait, no, in a \(45 - 45-90\) triangle, the legs are equal and the hypotenuse is \(leg\times\sqrt{2}\). Wait, I think I mixed up the sides. Let's define the right - triangle on the right: the right angle is between the height \(h\) and the base \(x\), and the angle at the bottom is \(45^{\circ}\). So, \(\tan(45^{\circ})=\frac{h}{x}\), but \(\tan(45^{\circ}) = 1\), so \(h = x\)? But that would mean \(x\approx4.2\), but that seems wrong. Wait, no, maybe the height \(h\) is the adjacent side and \(x\) is the opposite side. Wait, \(\tan(45^{\circ})=\frac{x}{h}\), and since \(\tan(45^{\circ}) = 1\), then \(x = h\). But let's check with another approach.

Wait, maybe the left triangle: we can also use the cosine function to find the base of the left triangle, but we don't need it. Wait, the key is that in the right - triangle on the right, angle is \(45^{\circ}\), so it's an isosceles right - triangle, so the two legs are equal. The height \(h\) is one leg, so the other leg (which is \(x\)) is equal to \(h\). But let's calculate \(h\) more precisely. \(h = 5\sin(57^{\circ})\approx5\times0.8386705679454239 = 4.19335\approx4.2\). Then, since the right - triangle on the right is isosceles (because angle is \(45^{\circ}\)), \(x = h\approx4.2\)? Wait, no, that can't be. Wait, I think I made a mistake in identifying the sides. Let's look at the diagram again. The big triangle is split into two right - triangles by the height. The left right - triangle has hypotenuse 5 and angle \(57^{\circ}\) at the top. The right right - triangle has angle \(45^{\circ}\) at the bottom. So the height \(h\) is common to both right - triangles.

In the left right - triangle: \(\sin(57^{\circ})=\frac{h}{5}\Rightarrow h = 5\sin(57^{\circ})\approx4.2\)

In the right right - triangle: \(\tan(45^{\circ})=\frac{h}{x}\)? No, \(\tan(45^{\circ})=\frac{x}{h}\)? Wait, no, in the right right - triangle, the angle at the bottom is \(45^{\circ}\), the right angle is between the height \(h\) and the base \(x\). So the side opposite to \(45^{\circ}\) is \(h\), and the side adjacent to \(45^{\circ}\) is \(x\). Then \(\tan(45^{\circ})=\frac{h}{x}\), and since \(\tan(45^{\circ}) = 1\), then \(h = x\). But that would mean \(x\approx4.2\). But let's check with \(\sin(45^{\circ})\). In the right right - triangle, \(\sin(45^{\circ})=\frac{h}{\text{hypotenuse}}\), but we don't know the hypotenuse. Wait, no, the side we want is \(x\), which is the adjacent side to the \(45^{\circ}\) angle. So \(\cos(45^{\circ})=\frac{x}{\text{hypotenuse}}\), but we don't know the hypotenuse. Alternatively, since it's a right - triangle with angle \(45^{\circ}\), the legs are equal, so \(x = h\).

Wait, maybe my initial approach is wrong. Let's do it again.

First, find the height \(h\) from the left triangle:

\(\sin(57^{\circ})=\frac{h}{5}\)

\(h = 5\times\sin(57^{\circ})\)

\(\sin(57^{\circ})\approx0.8387\), so \(h = 5\times0.8387 = 4.1935\approx4.2\)

Now, in the right triangle, angle is \(45^{\circ}\), and it's a right triangle, so the other non - right angle is also \(45^{\circ}\) (since the sum of angles in a triangle is \(180^{\circ}\), \(180 - 90 - 45=45^{\circ}\)). So it's an isosceles right triangle, so the two legs are equal. The height \(h\) is one leg, and \(x\) is the other leg. Therefore, \(x = h\approx4.2\)? But that seems too small. Wait, maybe I mixed up the angle. Wait, the angle at the top of the left triangle is \(57^{\circ}\), so the angle at the bottom of the left triangle is \(90 - 57 = 33^{\circ}\). The big triangle has angles: let's see, the two right - triangles are formed by the height. The right - triangle on the right has angle \(45^{\circ}\) at the bottom, so the angle at the top of the right - triangle is \(45^{\circ}\).

Wait, maybe the height \(h\) is equal to \(x\) because in a \(45 - 45-90\) triangle, legs are equal. So \(x = h\approx4.2\). But let's check with the cosine function in the left triangle. The base of the left triangle is \(b = 5\times\cos(57^{\circ})\approx5\times0.5446 = 2.723\approx2.7\). Then the total base of the big triangle is \(b + x\). But we don't know the total base. Wait, maybe the problem is that the right - triangle on the right is a \(45 - 45-90\) triangle, so \(x = h\). So \(x\approx4.2\). But let's calculate \(h\) more accurately.

\(\sin(57^{\circ})=\sin(57)\approx0.8386705679454239\)

\(h = 5\times0.8386705679454239 = 4.1933528397271195\approx4.2\)

So \(x = h\approx4.2\)? Wait, no, maybe I had the angle wrong. Wait, the angle in the right - triangle is \(45^{\circ}\), so \(\tan(45^{\circ}) = 1=\frac{h}{x}\), so \(x = h\). So \(x\approx4.2\).

Wait, but let's check with another method. Suppose we use the law of sines on the big triangle. The big triangle has angles: let's find the angles. The angle at the top of the big triangle is \(57^{\circ}+45^{\circ}=102^{\circ}\). The angle at the bottom left is \(33^{\circ}\) (since \(90 - 57 = 33\)), and the angle at the bottom right is \(45^{\circ}\). Wait, no, the big triangle: the two right - triangles are formed by the height, so the big triangle has angles: at the bottom left: \(90 - 57 = 33^{\circ}\), at the bottom right: \(45^{\circ}\), so the angle at the top is \(180-(33 + 45)=102^{\circ}\). The side opposite to \(33^{\circ}\) is \(x\)? No, the side opposite to \(33^{\circ}\) is the side with length 5? No, the side with length 5 is opposite to the angle at the bottom right (\(45^{\circ}\))? No, this is getting confusing.

Wait, let's go back to the two right - triangles.

Left right - triangle: hypotenuse = 5, angle = \(57^{\circ}\), right angle. So:

- Opposite to \(57^{\circ}\): \(h = 5\sin(57^{\circ})\approx4.2\)
- Adjacent to \(57^{\circ}\): \(a = 5\cos(57^{\circ})\approx5\times0.5446 = 2.723\approx2.7\)

Right right - triangle: right angle, angle = \(45^{\circ}\), leg = \(h\approx4.2\). Since it's a \(45 - 45-90\) triangle, the other leg \(x\) is equal to \(h\), so \(x\approx4.2\).

Answer:

\(4.2\)