Question

part 2 of 3
solve the triangle.
(image of triangle with side 5, side 6, angle 55° at the vertex between side 5 and side 6, side b opposite angle a, angle a at the vertex between side 6 and side b, angle c at the vertex between side 5 and side b)
b ≈ 5.16 (round to two decimal places as needed.)
a ≈ □° (round to one decimal place as needed.)
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question 25

Explanation:

Step1: Use Law of Sines

We know in a triangle, \(\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\). Here, we have side \(a = 5\), side \(c = 6\), angle \(B = 55^\circ\), and side \(b\approx5.16\). We can use \(\frac{\sin A}{a}=\frac{\sin B}{b}\). So \(\sin A=\frac{a\sin B}{b}\). Substitute \(a = 5\), \(\sin B=\sin55^\circ\approx0.8192\), \(b\approx5.16\). Then \(\sin A=\frac{5\times0.8192}{5.16}\approx\frac{4.096}{5.16}\approx0.7938\).

Step2: Find angle A

Take the inverse sine: \(A=\arcsin(0.7938)\approx52.5^\circ\) (we check the triangle sum, angle \(C = 180 - 55 - 52.5 = 72.5^\circ\), which is valid as angles sum to \(180^\circ\)).

Answer:

\(52.5\)