Question

part 1 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 c at the top vertex)
b ≈ \boxed{} (round to two decimal places as needed.)
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question 25

Explanation:

Step1: Identify the Law to Use

We have a triangle with two sides and the included angle? Wait, no, here we have side \( c = 5 \), side \( a = 6 \), angle \( A = 55^\circ \)? Wait, no, looking at the triangle: the side opposite angle \( A \) is \( 5 \)? Wait, no, the triangle has side 5, side 6, angle \( 55^\circ \) between side 5 and side 6? Wait, no, the labels: angle at the left is \( 55^\circ \), side adjacent to it is 5 (from angle to \( C \)) and side 6 (from angle to \( A \)), and side \( b \) is opposite the \( 55^\circ \) angle? Wait, no, let's use the Law of Cosines or Law of Sines. Wait, Law of Cosines: \( b^2 = 5^2 + 6^2 - 2 \times 5 \times 6 \times \cos(55^\circ) \)? Wait, no, if the angle is \( 55^\circ \), between sides 5 and 6, then yes, Law of Cosines.

Step2: Apply Law of Cosines

Law of Cosines formula: \( b^2 = c^2 + a^2 - 2ac \cos(B) \)? Wait, no, let's define the triangle: let's say the triangle has vertices: left vertex (angle \( 55^\circ \)), bottom side 6, left side 5, and side \( b \) opposite the \( 55^\circ \) angle? Wait, no, maybe Law of Sines. Wait, no, let's check the sides. Let's denote: angle \( A = 55^\circ \), side \( a = 6 \) (opposite angle \( A \)? No, wait, the side labeled 6 is the base, angle at the left is \( 55^\circ \), side from left angle to \( C \) is 5, so side \( AC = b \), side \( BC = 5 \), side \( AB = 6 \), angle at \( A \) is \( 55^\circ \)? Wait, no, the left angle is \( 55^\circ \), so angle at the left vertex (let's call it \( B \)) is \( 55^\circ \), side \( AB = 6 \), side \( BC = 5 \), and side \( AC = b \). Then we can use the Law of Cosines: \( b^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(angle B) \). So \( AB = 6 \), \( BC = 5 \), angle \( B = 55^\circ \). So \( b^2 = 6^2 + 5^2 - 2 \times 6 \times 5 \times \cos(55^\circ) \).

Step3: Calculate the Values

First, compute \( 6^2 = 36 \), \( 5^2 = 25 \), so \( 36 + 25 = 61 \). Then \( 2 \times 6 \times 5 = 60 \). \( \cos(55^\circ) \approx 0.5736 \). So \( 2 \times 6 \times 5 \times \cos(55^\circ) = 60 \times 0.5736 \approx 34.416 \). Then \( b^2 = 61 - 34.416 = 26.584 \). Then \( b = \sqrt{26.584} \approx 5.155 \), round to two decimal places: \( 5.16 \). Wait, no, wait, maybe I mixed up the angle. Wait, maybe the angle is not between the two sides. Wait, maybe it's Law of Sines. Wait, let's re-examine the triangle. The triangle has: angle at the left is \( 55^\circ \), side opposite to angle \( A \) (right angle) is 5? No, this is confusing. Wait, let's use Law of Cosines correctly. Let's define:

Let’s denote:

• Side \( a = 6 \) (opposite angle \( A \))

• Side \( c = 5 \) (opposite angle \( C \))

• Angle at the left (let's say angle \( B \)) is \( 55^\circ \)

Then, by Law of Cosines, \( b^2 = a^2 + c^2 - 2ac \cos(B) \)

So \( a = 6 \), \( c = 5 \), \( B = 55^\circ \)

So \( b^2 = 6^2 + 5^2 - 2 \times 6 \times 5 \times \cos(55^\circ) \)

Calculate each term:

\( 6^2 = 36 \)

\( 5^2 = 25 \)

\( 2 \times 6 \times 5 = 60 \)

\( \cos(55^\circ) \approx 0.5735764364 \)

So \( 2 \times 6 \times 5 \times \cos(55^\circ) = 60 \times 0.5735764364 \approx 34.41458618 \)

Then \( b^2 = 36 + 25 - 34.41458618 = 61 - 34.41458618 = 26.58541382 \)

Then \( b = \sqrt{26.58541382} \approx 5.1559 \), which rounds to \( 5.16 \)

Answer:

\( 5.16 \)