Question

find \\(\tan\theta\\), \\(csc\theta\\), and \\(cos\theta\\), where \\(\theta\\) is the angle shown in the figure. give exact values, not decimal approximations. figure of a right triangle with angle \\(\theta\\), one leg 5, hypotenuse 6, and the right angle marked

Explanation:

Step1: Identify triangle sides

In the right triangle, for angle \(\theta\):

• Opposite side (let's find the adjacent side first using Pythagoras). Hypotenuse \(= 6\), one leg \(= 5\). Let the other leg (adjacent to \(\theta\)) be \(a\). By Pythagoras: \(a^2 + 5^2 = 6^2\) → \(a^2 = 36 - 25 = 11\) → \(a = \sqrt{11}\). Wait, no, wait: Wait, the leg with length 5 is opposite or adjacent? Wait, the angle \(\theta\) is at the top, so the leg of length 5 is adjacent to \(\theta\)? Wait no, let's re - check. The right angle is at the bottom, so the sides: hypotenuse is 6, one leg (vertical) is 5, the other leg (horizontal) is \(x\). So \(x^2+5^2 = 6^2\) → \(x=\sqrt{36 - 25}=\sqrt{11}\). Wait, no, maybe I mixed up. Wait, \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\), \(\csc\theta=\frac{\text{hypotenuse}}{\text{opposite}}\), \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\).

Wait, let's define:

• Hypotenuse (\(h\)) = 6
• Opposite side to \(\theta\) (\(o\)): Let's see, the side opposite \(\theta\) is the horizontal leg. Wait, no, the vertical leg is 5, so if \(\theta\) is at the top, then the opposite side is the horizontal leg (length \(x\)), adjacent side is the vertical leg (length 5). So \(x=\sqrt{6^{2}-5^{2}}=\sqrt{36 - 25}=\sqrt{11}\).

Step2: Calculate \(\tan\theta\)

\(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}=\frac{\sqrt{11}}{5}\)

Step3: Calculate \(\csc\theta\)

\(\csc\theta=\frac{\text{hypotenuse}}{\text{opposite}}=\frac{6}{\sqrt{11}}=\frac{6\sqrt{11}}{11}\) (rationalizing the denominator)

Step4: Calculate \(\cos\theta\)

\(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{5}{6}\)

Wait, wait, I think I made a mistake. Wait, maybe the 5 is the opposite side. Let's re - assign. Let's say:

• Hypotenuse \(h = 6\)
• Opposite side to \(\theta\) (\(o\)) = 5
• Then adjacent side (\(a\)): \(a=\sqrt{h^{2}-o^{2}}=\sqrt{36 - 25}=\sqrt{11}\)

Then:

• \(\tan\theta=\frac{o}{a}=\frac{5}{\sqrt{11}}=\frac{5\sqrt{11}}{11}\)
• \(\csc\theta=\frac{h}{o}=\frac{6}{5}\)
• \(\cos\theta=\frac{a}{h}=\frac{\sqrt{11}}{6}\)

Ah, here's the mistake. I mixed up opposite and adjacent. If the right angle is at the bottom, and \(\theta\) is at the top, then the side of length 5 is opposite to \(\theta\), and the other leg (adjacent) is \(\sqrt{6^{2}-5^{2}}=\sqrt{11}\), hypotenuse is 6.

So:

For \(\tan\theta\):

\(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}=\frac{5}{\sqrt{11}}=\frac{5\sqrt{11}}{11}\) (rationalizing: multiply numerator and denominator by \(\sqrt{11}\))

For \(\csc\theta\):

\(\csc\theta=\frac{\text{hypotenuse}}{\text{opposite}}=\frac{6}{5}\)

For \(\cos\theta\):

\(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{\sqrt{11}}{6}\)

Answer:

\(\tan\theta=\boldsymbol{\frac{5\sqrt{11}}{11}}\), \(\csc\theta=\boldsymbol{\frac{6}{5}}\), \(\cos\theta=\boldsymbol{\frac{\sqrt{11}}{6}}\)