Question

suppose abc is a right triangle with sides of lengths a, b, and c and right angle at c. find the unknown side length using the pythagorean theorem and then find the value of the indicated trigonometric function of the given angle. rationalize the denominator if applicable. find tan(β) when b = 7 and c = 10. options: a. \\(\frac{\sqrt{51}}{7}\\), b. \\(\frac{10\sqrt{51}}{51}\\), c. \\(\frac{\sqrt{51}}{10}\\), d. \\(\frac{7\sqrt{51}}{51}\\)

Explanation:

Step1: Find side \( a \) using Pythagorean theorem

In a right triangle with right angle at \( C \), \( c \) is the hypotenuse, so \( a^2 + b^2 = c^2 \). Given \( b = 7 \), \( c = 10 \), we solve for \( a \):
\( a^2 = c^2 - b^2 = 10^2 - 7^2 = 100 - 49 = 51 \), so \( a = \sqrt{51} \).

Step2: Recall the definition of \( \tan(\beta) \)

In a right triangle, \( \tan(\beta) = \frac{\text{opposite}}{\text{adjacent}} \) to angle \( \beta \). For angle \( \beta \), the opposite side is \( a = \sqrt{51} \) and the adjacent side is \( b = 7 \). Wait, no—wait, angle \( \beta \): let's clarify the sides. Right angle at \( C \), so sides: \( a \) opposite \( A \), \( b \) opposite \( B \) (angle \( \beta \)), \( c \) hypotenuse. Wait, no—wait, \( \tan(\beta) \): angle \( \beta \) is at \( B \)? Wait, no, the triangle is \( ABC \) with right angle at \( C \), so vertices: \( C \) (right angle), \( A \) and \( B \). So side \( a \) is opposite \( A \) (BC), side \( b \) is opposite \( B \) (AC), hypotenuse \( c = AB \). So angle \( \beta \) is angle \( B \), so the opposite side to \( B \) is \( AC = b \)? Wait, no, I messed up. Wait, \( a \) is BC (opposite \( A \)), \( b \) is AC (opposite \( B \)), \( c \) is AB (hypotenuse). So angle \( \beta \) is angle \( B \), so for angle \( B \), the opposite side is \( AC = b \)? No, wait, no: in triangle \( ABC \), right-angled at \( C \), so:
- \( \angle C = 90^\circ \)
- \( \angle A \) and \( \angle B \) are acute.
- Side opposite \( \angle A \) is \( BC = a \)
- Side opposite \( \angle B \) is \( AC = b \)
- Hypotenuse \( AB = c \)

Wait, the problem says "Find \( \tan(\beta) \) when \( b = 7 \) and \( c = 10 \)". Let's assume \( \beta \) is \( \angle B \). Then, \( \tan(\beta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{AC}{BC} = \frac{b}{a} \)? Wait, no—wait, \( \angle B \): the sides adjacent to \( \angle B \) is \( BC = a \), and opposite is \( AC = b \)? Wait, no, adjacent to \( \angle B \) is \( AB \)? No, no. Let's draw the triangle: \( C \) is right angle, so \( AC \) and \( BC \) are legs, \( AB \) is hypotenuse. So \( AC = b \), \( BC = a \), \( AB = c \). Then \( \angle B \) is at vertex \( B \), so the sides:
- Opposite \( \angle B \): \( AC = b \)
- Adjacent \( \angle B \): \( BC = a \)

Wait, but we found \( a = \sqrt{51} \), \( b = 7 \). Wait, no—wait, \( a^2 + b^2 = c^2 \), so \( a = \sqrt{c^2 - b^2} = \sqrt{100 - 49} = \sqrt{51} \). So \( a = \sqrt{51} \) (BC), \( b = 7 \) (AC), \( c = 10 \) (AB). Then \( \tan(\beta) \), where \( \beta = \angle B \):
\( \tan(\beta) = \frac{\text{opposite to } \beta}{\text{adjacent to } \beta} = \frac{AC}{BC} = \frac{b}{a} = \frac{7}{\sqrt{51}} \). But we need to rationalize the denominator: multiply numerator and denominator by \( \sqrt{51} \):
\( \frac{7\sqrt{51}}{51} \). Wait, but let's check the options. Option D is \( \frac{7\sqrt{51}}{51} \)? Wait, no—wait, maybe I mixed up opposite and adjacent. Wait, maybe \( \beta \) is \( \angle A \)? Wait, no, let's re-express.

Wait, maybe the angle \( \beta \) is such that the opposite side is \( a \) and adjacent is \( b \). Wait, let's re-express the tangent definition. In a right triangle, \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \). Let's confirm the sides:

- Hypotenuse \( c = 10 \)
- One leg \( b = 7 \) (let's say \( AC = 7 \))
- The other leg \( a = \sqrt{51} \) (BC = \( \sqrt{51} \))

If \( \beta \) is \( \angle A \), then opposite side is \( BC = \sqrt{51} \), adjacent side is \( AC = 7 \), so \( \tan(\beta) = \frac{\sqrt{51}}{7} \)? But that's option A. Wait, this is confusing. Wait, maybe the problem has \( \beta \) as the angle opposite to side \( a \). Wait, let's re-express the Pythagorean theorem: \( a^2 + b^2 = c^2 \), so \( a = \sqrt{c^2 - b^2} = \sqrt{100 - 49} = \sqrt{51} \). Then, \( \tan(\beta) \): if \( \beta \) is the angle with opposite side \( a \) and adjacent side \( b \), then \( \tan(\beta) = \frac{a}{b} = \frac{\sqrt{51}}{7} \)? But that's option A. Wait, no—wait, maybe I made a mistake in the side labels. Let's check the options. Option D is \( \frac{7\sqrt{51}}{51} \), which is \( \frac{7}{\sqrt{51}} \) rationalized. Wait, maybe \( \beta \) is the angle where the adjacent side is \( a \) and opposite is \( b \). Wait, let's re-express:

Wait, the problem says "Find \( \tan(\beta) \) when \( b = 7 \) and \( c = 10 \)". Let's assume that \( b \) is the adjacent side, and \( a \) is the opposite side. Wait, no—let's do it step by step.

1. Find \( a \): \( a = \sqrt{c^2 - b^2} = \sqrt{10^2 - 7^2} = \sqrt{100 - 49} = \sqrt{51} \).

2. Recall that \( \tan(\beta) = \frac{\text{opposite}}{\text{adjacent}} \). Now, we need to determine which side is opposite and which is adjacent to \( \beta \). Let's assume \( \beta \) is the angle at \( B \), so:

- Opposite side to \( \beta \): \( AC = b = 7 \)
- Adjacent side to \( \beta \): \( BC = a = \sqrt{51} \)

Thus, \( \tan(\beta) = \frac{7}{\sqrt{51}} \). Rationalizing the denominator: multiply numerator and denominator by \( \sqrt{51} \):

\( \tan(\beta) = \frac{7\sqrt{51}}{51} \), which is option D. Wait, but let's check the options again. Option D is \( \frac{7\sqrt{51}}{51} \), yes. So that's the answer.

Answer:

D. \(\frac{7\sqrt{51}}{51}\)