Question

use the triangle shown to find the ratios. cos(a) = tan(c) =

Explanation:

For \(\boldsymbol{\cos(A)}\):

Step1: Recall cosine definition

In a right triangle, \(\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}\) for angle \(\theta\). For \(\angle A\), adjacent side to \(A\) is \(AB = 10\) ft, hypotenuse is \(AC = 26\) ft? Wait, no, wait. Wait, in right triangle \(ABC\) with right angle at \(B\), so sides: \(AB = 10\) (opposite \(C\)), \(BC = 24\) (adjacent to \(A\) and opposite \(A\)? Wait, no. Let's label: right angle at \(B\), so:

• For \(\angle A\):
• Adjacent side: \(AB\)? Wait, no. Wait, in angle \(A\), the sides:
• Opposite: \(BC = 24\) ft
• Adjacent: \(AB = 10\) ft? Wait, no, hypotenuse is \(AC = 26\) ft. Wait, no, cosine of angle \(A\) is adjacent over hypotenuse. Wait, angle \(A\) is at vertex \(A\), so the sides:
• The two legs: \(AB\) (length 10) and \(BC\) (length 24)
• Hypotenuse: \(AC\) (length 26)
• For angle \(A\):
• Adjacent side: \(AB\) (since it's one of the legs forming angle \(A\))
• Opposite side: \(BC\) (the leg opposite angle \(A\))
• Hypotenuse: \(AC\)

So \(\cos(A)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{AB}{AC}=\frac{10}{26}=\frac{5}{13}\)? Wait, no, wait, maybe I mixed up. Wait, angle \(A\) is between \(AB\) and \(AC\). So the adjacent side to angle \(A\) is \(AB\) (length 10), hypotenuse is \(AC\) (length 26). Wait, but let's check with Pythagoras: \(10^2 + 24^2 = 100 + 576 = 676 = 26^2\), so that's correct. So \(\cos(A)=\frac{\text{adjacent to }A}{\text{hypotenuse}}=\frac{AB}{AC}=\frac{10}{26}=\frac{5}{13}\)? Wait, no, wait, maybe I got adjacent and opposite wrong. Wait, angle \(A\): the sides:

• The side adjacent to \(A\) is \(AB\) (since it's one leg connected to \(A\))
• The side opposite to \(A\) is \(BC\) (the other leg, not connected to \(A\))
• Hypotenuse is \(AC\)

So \(\cos(A)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{AB}{AC}=\frac{10}{26}=\frac{5}{13}\). Wait, but let's confirm with the definition: in a right triangle, \(\cos(\theta)=\frac{\text{length of adjacent side}}{\text{length of hypotenuse}}\). So for angle \(A\), adjacent side is \(AB\) (10 ft), hypotenuse is \(AC\) (26 ft). So \(\cos(A)=\frac{10}{26}=\frac{5}{13}\). Wait, but maybe I made a mistake. Wait, alternatively, maybe adjacent is \(BC\)? No, angle \(A\) is at \(A\), so the two sides forming angle \(A\) are \(AB\) and \(AC\). So \(AB\) is adjacent, \(AC\) is hypotenuse. So yes, \(\cos(A)=\frac{AB}{AC}=\frac{10}{26}=\frac{5}{13}\).

Step2: Calculate \(\cos(A)\)

Given \(AB = 10\) ft (adjacent to \(A\)), \(AC = 26\) ft (hypotenuse). So:

$$ \cos(A) = \frac{\text{adjacent to } A}{\text{hypotenuse}} = \frac{AB}{AC} = \frac{10}{26} = \frac{5}{13} $$

For \(\boldsymbol{\tan(C)}\):

Step1: Recall tangent definition

In a right triangle, \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}\) for angle \(\theta\). For \(\angle C\):

• Opposite side: \(AB = 10\) ft (the leg opposite angle \(C\))
• Adjacent side: \(BC = 24\) ft (the leg adjacent to angle \(C\))

So \(\tan(C)=\frac{\text{opposite to }C}{\text{adjacent to }C}=\frac{AB}{BC}\)

Step2: Calculate \(\tan(C)\)

Given \(AB = 10\) ft (opposite \(C\)), \(BC = 24\) ft (adjacent \(C\)):

$$ \tan(C) = \frac{\text{opposite to } C}{\text{adjacent to } C} = \frac{AB}{BC} = \frac{10}{24} = \frac{5}{12} $$

Answer:

s:

• \(\cos(A) = \boldsymbol{\frac{5}{13}}\)
• \(\tan(C) = \boldsymbol{\frac{5}{12}}\)