Question

1. a right triangle with legs 5 and 7, and angle x° opposite the leg of length 5 (or adjacent? wait, the right angle is at the left, so the sides: one leg is 5, another leg is 7, and we need to find angle x°).

Explanation:

Step1: Identify the triangle type

This is a right - triangle with legs of length 5 and 7, and we want to find the angle \(x^{\circ}\) opposite the side of length 5 (wait, no, actually, let's correct: the side with length 5 and 7 are the legs. To find angle \(x\), we can use trigonometric ratios. Let's recall that in a right - triangle, \(\tan(x)=\frac{\text{opposite}}{\text{adjacent}}\). Wait, let's see the sides: the side of length 5 and 7 are the two legs. Let's assume that the angle \(x\) has the opposite side of length 5 and adjacent side of length 7? Wait, no, maybe we should use the tangent function. Wait, \(\tan(x)=\frac{5}{7}\) or \(\tan(x)=\frac{7}{5}\)? Wait, let's look at the triangle: the right - angle is at the vertex where the sides of length 5 and 7 meet. So the angle \(x\) is between the side of length 5 and the hypotenuse? Wait, no, let's label the triangle: let's call the right - angle \(C\), the vertex with angle \(x\) as \(A\), and the other vertex as \(B\). Then side \(AC = 5\), side \(BC=7\), and right - angle at \(C\). Then angle \(x\) is at \(A\), so \(\tan(x)=\frac{BC}{AC}=\frac{7}{5}\)? Wait, no, \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}\). For angle \(A\) (angle \(x\)), the opposite side is \(BC = 7\) and adjacent side is \(AC = 5\)? Wait, no, \(AC = 5\) is adjacent to angle \(A\), and \(BC = 7\) is opposite to angle \(A\). So \(\tan(x)=\frac{7}{5}\)? Wait, no, maybe I got it wrong. Wait, let's use the tangent function correctly. If we have a right - triangle, and we want to find an acute angle \(x\), we can use \(\tan(x)=\frac{\text{opposite}}{\text{adjacent}}\). Let's suppose that the side of length 5 is adjacent to angle \(x\) and the side of length 7 is opposite to angle \(x\). Then \(\tan(x)=\frac{7}{5}\)? Wait, no, maybe it's \(\tan(x)=\frac{5}{7}\). Wait, let's calculate \(\tan^{-1}(\frac{5}{7})\) or \(\tan^{-1}(\frac{7}{5})\). Wait, let's compute \(\tan^{-1}(\frac{5}{7})\): \(\frac{5}{7}\approx0.714\), and \(\tan^{-1}(0.714)\approx35.54^{\circ}\). If we compute \(\tan^{-1}(\frac{7}{5})=\tan^{-1}(1.4)\approx54.46^{\circ}\). Wait, maybe we should use the correct ratio. Let's re - examine the triangle: the two legs are 5 and 7. So to find angle \(x\), let's use the tangent function. Let's assume that the angle \(x\) is such that the opposite side is 5 and adjacent side is 7. Then \(\tan(x)=\frac{5}{7}\). Then \(x=\tan^{-1}(\frac{5}{7})\). Let's calculate that: \(\frac{5}{7}\approx0.7143\), and \(\tan^{-1}(0.7143)\approx35.54^{\circ}\).

Wait, maybe I made a mistake in identifying the opposite and adjacent sides. Let's use the definition: in a right - triangle, for an acute angle \(x\), \(\sin(x)=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos(x)=\frac{\text{adjacent}}{\text{hypotenuse}}\), \(\tan(x)=\frac{\text{opposite}}{\text{adjacent}}\). First, we need to find the hypotenuse. By the Pythagorean theorem, \(c=\sqrt{5^{2}+7^{2}}=\sqrt{25 + 49}=\sqrt{74}\approx8.602\). Then, if we want to find angle \(x\), let's say the side opposite to \(x\) is 5, then \(\sin(x)=\frac{5}{\sqrt{74}}\), and \(x=\sin^{-1}(\frac{5}{\sqrt{74}})\approx\sin^{-1}(\frac{5}{8.602})\approx\sin^{-1}(0.581)\approx35.54^{\circ}\). If the side opposite to \(x\) is 7, then \(\sin(x)=\frac{7}{\sqrt{74}}\approx\sin^{-1}(\frac{7}{8.602})\approx\sin^{-1}(0.814)\approx54.46^{\circ}\). But from the diagram, the side of length 5 is shorter than the side of length 7, so the angle opposite to the shorter side (5) should be smaller. So angle \(x\) is opposite to the side of length 5? Wait, no, in the diagram, the side of length 5 and 7 are the legs. The angle \(x\) is between the side of length 5 and the hypotenuse. Wait, maybe the correct approach is:

Let's use the tangent function. Let's assume that angle \(x\) has the opposite side of length 5 and adjacent side of length 7. Then \(\tan(x)=\frac{5}{7}\). Then \(x=\arctan(\frac{5}{7})\). Let's calculate \(\frac{5}{7}\approx0.714\). Using a calculator, \(\arctan(0.714)\approx35.5^{\circ}\).

Step1: Recall the trigonometric ratio for tangent

In a right - triangle, \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}\), where \(\theta\) is an acute angle, "opposite" is the side opposite to \(\theta\), and "adjacent" is the side adjacent to \(\theta\) (other than the hypotenuse).

Step2: Identify opposite and adjacent sides for angle \(x\)

In the given right - triangle, the two legs are of lengths 5 and 7. Let the angle \(x\) have the side of length 5 as the opposite side and the side of length 7 as the adjacent side. So \(\tan(x)=\frac{5}{7}\).

Step3: Calculate the value of \(x\)

We know that \(x=\arctan(\frac{5}{7})\). Calculate \(\frac{5}{7}\approx0.714\). Using a calculator, \(\arctan(0.714)\approx35.5^{\circ}\) (or more precisely, \(\arctan(\frac{5}{7})\approx35.5377^{\circ}\)).

Answer:

\(x\approx35.5^{\circ}\) (or \(x = \arctan(\frac{5}{7})\approx35.5^{\circ}\))