Problem 37 - College Trigonometry

Question

Given \(\alpha=30^{\circ}\) and \(b=10\), choose four different values for \(a\) so that (a) the information yields no triangle (b) the information yields exactly one right triangle (c) the information yields two distinct triangles (d) the information yields exactly one obtuse triangle Explain why you cannot choose \(a\) in such a way as to have \(\alpha=30^{\circ}, b=10\) and your choice of \(a\) yield only one triangle where that unique triangle has three acute angles.

Step-by-Step Solution

Answer

a<5: no triangle, a=5: one right triangle, 5

Step 1: Understanding the Information Given

We are given an angle \(\alpha = 30^{\circ}\) and a side \(b = 10\). We need to use the law of sines to find whether different values of \(a\) can form triangles.

Step 2: Using the Law of Sines

The Law of Sines states that \( \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} \). Given \( A = 30^{\circ} \) and \( b = 10 \), we express \(a = 20 \cdot \sin(B)\). Knowing \(\sin(30^{\circ}) = 0.5\), decide how \(a\) affects the possibility of forming triangles.

Step 3: Condition for No Triangle

No triangle is formed when the length of side \(a\) is too short. Calculate the perpendicular from angle \(30^{\circ}\) to side \(b\) as height \(h = b \cdot \sin(30^{\circ}) = 10 \times 0.5 = 5\). If \(a < 5\), no triangle can be formed.

Step 4: Condition for Exactly One Right Triangle

A right triangle is formed if \(a\) equals the height \(h\). Substitute \(a = 5\), meaning the triangle has an angle \(sin^{-1}(5/10) = 30^{\circ}\), providing only one such right triangle.

Step 5: Condition for Two Distinct Triangles

Two triangles can be formed if \(a\) is more than the height \(h\) but less than the length of \(b = 10\). This is only possible when \(5 < a < 10\). Choosing \(a = 8\), we can draw two triangles satisfying this inequality.

Step 6: Condition for Exactly One Obtuse Triangle

An obtuse triangle occurs if \(a \geq b = 10\). If \(a = 15\), the longest side constraint ensures an obtuse triangle where angle opposite \(a\) exceeds \(90^{\circ}\).

Step 7: Explaining the Absence of All Acute Triangles

For an acute triangle, the Law of Sines demands that each corresponding angle be less than \(90^{\circ}\). When \(b = 10\) we cannot force the remaining two angles to both be less than \(90^{\circ}\), as demonstrated by constraints and the sum of angles.

Key Concepts

Understanding Triangle Conditions

When exploring the possibilities of forming triangles given specific conditions, it's crucial to consider the triangle inequality theorem and the law of sines. In this exercise, we are working with the angle \( \alpha = 30^{\circ} \) and a side \( b = 10 \). This sets the stage to explore different possible scenarios by varying the length of side \( a \).

• The triangle inequality theorem states that for any triangle, the sum of the lengths of any two sides must be greater than the length of the remaining side.
• Given our scenario, for a valid triangle to form, the length of \( a \) in conjunction with \( b \) must respect this rule.

The length of \( a \) affects whether a triangle can be formed and determines the type of triangle (right, obtuse, or multiple possible triangles). Understanding these conditions helps in planning which values to explore.

Right Triangle Identification

Identifying a right triangle involves finding when one angle of the triangle is exactly \( 90^{\circ} \). In our scenario with \( \alpha = 30^{\circ} \) and given \( b = 10 \):

• A right triangle is formed if the length of \( a \) is exactly equal to the height formed by drawing a perpendicular from angle \( 30^{\circ} \) to side \( b \).
• This perpendicular, or height, can be calculated as \( h = b \cdot \sin(30^{\circ}) = 5 \).

Therefore, a right triangle is formed when \( a = 5 \). This choice results in a right triangle where the right angle is at \( \text{B} \). It is crucial in this context because it provides a clear distinct formation.

Obtuse Triangle Formation

Forming an obtuse triangle implies having one angle greater than \( 90^{\circ} \). With \( a \), \( b \), and \( \alpha = 30^{\circ} \):

• An obtuse triangle is formed when the side opposite the angle is the longest side among the three sides of the triangle, meaning \( a \geq b \).
• For our given \( b = 10 \), having \( a > 10 \) guarantees that the opposite angle forming with side \( a \) will be greater than \( 90^{\circ} \).

Choosing \( a = 15 \) ensures that the triangle will be obtuse, as according to the triangle inequality and side conditions, \( a \) being longer guarantees the obtuse configuration of the angles.

Explaining Acute Triangle Constraints

Understanding why an acute triangle solely cannot be formed with our given conditions of \( \alpha = 30^{\circ} \) and \( b = 10 \) requires pondering over angle possibilities:

• An acute triangle is characterized by having all angles less than \( 90^{\circ} \).
• In our case, forming a triangle where side \( b = 10 \) is among the longest while trying to keep all angles less than \( 90^{\circ} \) becomes impossible.

The issue stems from the impossibility of configuring the remainder of the angles, alongside \( 30^{\circ} \), in such a way that each remains acute given the fixed side \( b \) and adjustable \( a \). Hence, an all-acutely nonetheless becomes a result inaccessible within our stipulated parameters.

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Problem 38

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Key Concepts

Understanding Triangle Conditions

When exploring the possibilities of forming triangles given specific conditions, it's crucial to consider the triangle inequality theorem and the law of sines. In this exercise, we are working with the angle \( \alpha = 30^{\circ} \) and a side \( b = 10 \). This sets the stage to explore different possible scenarios by varying the length of side \( a \).

• The triangle inequality theorem states that for any triangle, the sum of the lengths of any two sides must be greater than the length of the remaining side.
• Given our scenario, for a valid triangle to form, the length of \( a \) in conjunction with \( b \) must respect this rule.

The length of \( a \) affects whether a triangle can be formed and determines the type of triangle (right, obtuse, or multiple possible triangles). Understanding these conditions helps in planning which values to explore.

Right Triangle Identification

Identifying a right triangle involves finding when one angle of the triangle is exactly \( 90^{\circ} \). In our scenario with \( \alpha = 30^{\circ} \) and given \( b = 10 \):

• A right triangle is formed if the length of \( a \) is exactly equal to the height formed by drawing a perpendicular from angle \( 30^{\circ} \) to side \( b \).
• This perpendicular, or height, can be calculated as \( h = b \cdot \sin(30^{\circ}) = 5 \).

Therefore, a right triangle is formed when \( a = 5 \). This choice results in a right triangle where the right angle is at \( \text{B} \). It is crucial in this context because it provides a clear distinct formation.

Obtuse Triangle Formation

Forming an obtuse triangle implies having one angle greater than \( 90^{\circ} \). With \( a \), \( b \), and \( \alpha = 30^{\circ} \):

• An obtuse triangle is formed when the side opposite the angle is the longest side among the three sides of the triangle, meaning \( a \geq b \).
• For our given \( b = 10 \), having \( a > 10 \) guarantees that the opposite angle forming with side \( a \) will be greater than \( 90^{\circ} \).

Choosing \( a = 15 \) ensures that the triangle will be obtuse, as according to the triangle inequality and side conditions, \( a \) being longer guarantees the obtuse configuration of the angles.

Explaining Acute Triangle Constraints

Understanding why an acute triangle solely cannot be formed with our given conditions of \( \alpha = 30^{\circ} \) and \( b = 10 \) requires pondering over angle possibilities:

• An acute triangle is characterized by having all angles less than \( 90^{\circ} \).
• In our case, forming a triangle where side \( b = 10 \) is among the longest while trying to keep all angles less than \( 90^{\circ} \) becomes impossible.

The issue stems from the impossibility of configuring the remainder of the angles, alongside \( 30^{\circ} \), in such a way that each remains acute given the fixed side \( b \) and adjustable \( a \). Hence, an all-acutely nonetheless becomes a result inaccessible within our stipulated parameters.