Question

part 3 of 3
solve the triangle.
(image of a triangle with side 5, side 6, angle 55° at the vertex between side 5 and side 6, side b opposite angle a, angle a opposite side 6, angle c opposite side 6? wait, no, the triangle has vertices: lets see, the triangle has a side of length 5, a side of length 6, angle 55° between the side of length 5 and the side of length 6? wait, the image shows a triangle with vertex c, side 5 from c to the left vertex, side 6 from the left vertex to vertex a, side b from c to a. the angle at the left vertex (between side 5 and side 6) is 55°. so we have triangle with sides: one side 5, one side 6, included angle 55°? wait, no, the angle is 55° at the vertex with side 5 and side 6? wait, the ocr text:
b ≈ 5.16 (round to two decimal places as needed.)
a ≈ 52.5 ° (round to one decimal place as needed.)
c ≈ □° (round to one decimal place as needed.)
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question 25 of 35

Explanation:

Step1: Recall triangle angle sum

The sum of angles in a triangle is \(180^\circ\). Let the given angle be \(55^\circ\) (let's call it angle opposite side 6? Wait, no, the given angle is \(55^\circ\), angle \(A \approx 52.5^\circ\), so we need to find angle \(C\).

Step2: Calculate angle \(C\)

Using the formula \(A + B + C = 180^\circ\), here \(B = 55^\circ\), \(A \approx 52.5^\circ\), so \(C = 180 - 55 - 52.5\).
Calculate \(180 - 55 = 125\), then \(125 - 52.5 = 72.5\). So \(C \approx 72.5^\circ\).

Answer:

\(72.5\)