Determine the number of triangles ABC possible with the given parts.
a = 31, b = 26, B = 48°
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Identify the given elements: side \(a = 31\), side \(b = 26\), and angle \(B = 48^\circ\). Note that angle \(B\) is opposite side \(b\).
Use the Law of Sines to find the possible values of angle \(A\). The Law of Sines states: \(\frac{a}{\sin A} = \frac{b}{\sin B}\). Substitute the known values to get \(\frac{31}{\sin A} = \frac{26}{\sin 48^\circ}\).
Rearrange the equation to solve for \(\sin A\): \(\sin A = \frac{31 \times \sin 48^\circ}{26}\). Calculate the right side to find the value of \(\sin A\) (do not compute the final value here).
Analyze the value of \(\sin A\) to determine the number of possible triangles: if \(\sin A > 1\), no triangle exists; if \(\sin A = 1\), exactly one right triangle exists; if \(0 < \sin A < 1\), there may be one or two possible triangles depending on the angle \(A\) and the sum of angles.
If two triangles are possible, find the second possible angle \(A' = 180^\circ - A\). Then check if the sum of angles \(A' + B\) is less than \(180^\circ\) to confirm the existence of the second triangle.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Law of Sines
The Law of Sines relates the sides and angles of a triangle through the ratio a/sin(A) = b/sin(B) = c/sin(C). It is essential for solving triangles when given two sides and an angle not included between them, allowing calculation of unknown angles or sides.
When two sides and a non-included angle (SSA) are given, there can be zero, one, or two possible triangles. This ambiguity arises because the given angle and side lengths may correspond to different triangle configurations, requiring careful analysis to determine the number of valid solutions.
The triangle inequality states that the sum of any two sides must be greater than the third side. Additionally, angles must sum to 180°. These constraints help verify if the given measurements can form a valid triangle and limit the possible number of triangles.