Question

7.1 hw - right triangle
etry; applications
question 8, 7.1.41
part 1 of 3
hw score: 38.46%, 5 of 13 points
points: 0 of 1
st
use the right triangle and the given information to solve the triangle
a = 1, c = 2, find b, a, and b
b = (round to the nearest hundredth as needed.)

Explanation:

Step1: Use Pythagorean theorem

In a right - triangle, \(a^{2}+b^{2}=c^{2}\). Given \(a = 1\) and \(c = 2\), we can solve for \(b\) as \(b=\sqrt{c^{2}-a^{2}}\).
Substitute \(a = 1\) and \(c = 2\) into the formula: \(b=\sqrt{2^{2}-1^{2}}=\sqrt{4 - 1}=\sqrt{3}\approx1.73\)

Step2: Find angle \(A\)

We know that \(\sin A=\frac{a}{c}\). Substitute \(a = 1\) and \(c = 2\) into the formula: \(\sin A=\frac{1}{2}\). So \(A = 30^{\circ}\)

Step3: Find angle \(B\)

Since \(A + B=90^{\circ}\) (in a right - triangle), then \(B = 90^{\circ}-A\). Substitute \(A = 30^{\circ}\), we get \(B = 60^{\circ}\)

Answer:

\(b\approx1.73\), \(A = 30^{\circ}\), \(B = 60^{\circ}\)