Question

use the triangle shown on the right and the given information to solve the triangle
a = 9, a = 25°, find b, c, and b

b = \boxed{} (round to two decimal places as needed )

Explanation:

Step1: Identify the triangle type

Assume it's a right - triangle (since we can solve for sides and angles with one side and one acute angle). In a right - triangle, we know that \(A + B=90^{\circ}\) (because the sum of acute angles in a right - triangle is \(90^{\circ}\)) and \(\tan(A)=\frac{a}{b}\) (opposite over adjacent in a right - triangle, where \(a\) is opposite angle \(A\) and \(b\) is adjacent to angle \(A\)).

Step2: Find angle \(B\)

Since \(A + B = 90^{\circ}\) and \(A = 25^{\circ}\), then \(B=90^{\circ}- 25^{\circ}=65^{\circ}\)

Step3: Find side \(b\)

We know that \(\tan(A)=\frac{a}{b}\), so \(b=\frac{a}{\tan(A)}\)
Given \(a = 9\) and \(A=25^{\circ}\), \(\tan(25^{\circ})\approx0.4663\)
Then \(b=\frac{9}{0.4663}\approx19.30\)

Step4: Find side \(c\) (hypotenuse)

We can use the cosine function: \(\cos(A)=\frac{b}{c}\), so \(c=\frac{b}{\cos(A)}\)
We know \(b\approx19.30\) and \(\cos(25^{\circ})\approx0.9063\)
\(c=\frac{19.30}{0.9063}\approx21.29\) (or we can use the sine function \(\sin(A)=\frac{a}{c}\), so \(c = \frac{a}{\sin(A)}\), \(\sin(25^{\circ})\approx0.4226\), \(c=\frac{9}{0.4226}\approx21.29\))

Answer:

\(b\approx19.30\), \(B = 65^{\circ}\), \(c\approx21.29\) (If we only consider the value of \(b\) as asked in the box first, \(b\approx19.30\))