Question

consider a triangle \\(abc\\) like the one below. suppose that \\(b = 52^{\circ}\\), \\(a = 69\\), and \\(c = 56\\). (the figure is not drawn to scale.) solve the triangle.

carry your intermediate computations to at least four decimal places, and round your answers to the nearest tenth.

if there is more than one solution, use the button labeled \or\.

\\(a = \square^{\circ}\\), \\(c = \square^{\circ}\\), \\(b = \square\\)

Explanation:

Identify given values and target unknowns

We are given a triangle \(ABC\) with:

• Angle \(B = 52^\circ\)
• Side \(a = 69\)
• Side \(c = 56\)

We need to solve the triangle, which means finding:

• Side \(b\)
• Angle \(A\)
• Angle \(C\)

Apply the Law of Cosines to find side b

To find side \(b\) when we know sides \(a\), \(c\), and the included angle \(B\), we use the Law of Cosines:

$$b^2 = a^2 + c^2 - 2ac \cos(B)$$

Substitute the given values:

$$b^2 = 69^2 + 56^2 - 2(69)(56) \cos(52^\circ)$$

$$b^2 = 4761 + 3136 - 7728 \cos(52^\circ)$$

$$b^2 \approx 7897 - 7728(0.615661)$$

$$b^2 \approx 7897 - 4757.8319$$

$$b^2 \approx 3139.1681$$

$$b \approx \sqrt{3139.1681} \approx 56.0283$$

Rounding to the nearest tenth:

$$b \approx 56.0$$

Apply the Law of Sines to find angle A

Now we use the Law of Sines to find one of the remaining angles. Let's find angle \(A\):

$$\frac{\sin(A)}{a} = \frac{\sin(B)}{b}$$

$$\sin(A) = \frac{a \sin(B)}{b}$$

Substitute the values (using high-precision intermediate values):

$$\sin(A) = \frac{69 \sin(52^\circ)}{56.0283}$$

$$\sin(A) \approx \frac{69(0.788011)}{56.0283}$$

$$\sin(A) \approx \frac{54.3727}{56.0283} \approx 0.970451$$

Find the angle \(A\):

$$A = \sin^{-1}(0.970451) \approx 76.03^\circ$$

Since \(a > c\), angle \(A\) must be larger than angle \(C\). Let's check if \(A\) could be obtuse:
If \(A \approx 180^\circ - 76.03^\circ = 103.97^\circ\), then \(A + B = 103.97^\circ + 52^\circ = 155.97^\circ < 180^\circ\), which is geometrically possible.
However, since \(b \approx 56.0283\) is very close to \(c = 56\), angle \(B\) and angle \(C\) must be very close in measure.
Let's calculate angle \(C\) directly using the Law of Sines to resolve any ambiguity:

$$\sin(C) = \frac{c \sin(B)}{b} = \frac{56 \sin(52^\circ)}{56.0283} \approx \frac{56(0.788011)}{56.0283} \approx 0.787613$$

$$C = \sin^{-1}(0.787613) \approx 51.96^\circ$$

Since \(c < b\) (\(56 < 56.0283\)), angle \(C\) must be slightly smaller than angle \(B\) (\(52^\circ\)), which matches \(51.96^\circ\).
Thus, angle \(C\) is acute:

$$C \approx 52.0^\circ$$

Calculate angle A using the angle sum of a triangle

Now we find angle \(A\) using the sum of angles in a triangle:

$$A = 180^\circ - B - C$$

$$A \approx 180^\circ - 52^\circ - 51.96^\circ = 76.04^\circ$$

Rounding to the nearest tenth:

• \(A \approx 76.0^\circ\)
• \(C \approx 52.0^\circ\)
• \(b \approx 56.0\)

Answer:

Consider a triangle \(ABC\) like the one below. Suppose that \(B = 52^\circ\), \(a = 69\), and \(c = 56\). Solve the triangle.

\(A =\) <blank>\(76.0\)</blank>\(^\circ\), \(C =\) <blank>\(52.0\)</blank>\(^\circ\), \(b =\) <blank>\(56.0\)</blank>