Question

draw triangles by sides & angles
two sides of a triangle are 5 cm long. one angle measure is known.
if that angle measures 36°, dropdown triangle(s) can be made.
if that angle measures 90°, dropdown triangle(s) can be made.
if that angle measures 144°, dropdown triangle(s) can be made.

Explanation:

Step1: Analyze 36° case

We have two sides of 5 cm. Let's consider the angle:
- If the 36° angle is between the two 5 - cm sides (included angle), we can form one triangle by SAS (Side - Angle - Side) criterion.
- If the 36° angle is not between the two 5 - cm sides (i.e., it is an angle opposite one of the 5 - cm sides), we can also form another triangle. So, for 36°, we can form 2 triangles? Wait, no, wait. Wait, when two sides are equal (5 cm each), so it's an isosceles triangle situation.
Wait, let's re - think. The two sides are 5 cm. Let's denote the triangle as \( \triangle ABC \) with \( AB = AC=5\) cm.
- Case 1: The known angle is \( \angle A = 36^{\circ}\) (included angle between \( AB \) and \( AC \)). Then, by SAS, we can form one triangle.
- Case 2: The known angle is \( \angle B=36^{\circ}\). Since \( AB = AC \), \( \angle B=\angle C = 36^{\circ}\), and we can find \( \angle A=180 - 2\times36=108^{\circ}\), so we can form another triangle. Wait, no, actually, when we have two equal sides, if the given angle is not the included angle, we have to check the triangle inequality and the angle - side relationship.
Wait, the sum of angles in a triangle is \( 180^{\circ}\). For a triangle with two sides of length \( a = b = 5\) cm.
If the angle \( \theta=36^{\circ}\):
- If \( \theta \) is the vertex angle (between the two equal sides), then the base angles are \( \frac{180 - 36}{2}=72^{\circ}\), and we can form one triangle.
- If \( \theta \) is a base angle, then the vertex angle is \( 180-2\times36 = 108^{\circ}\), and we can form another triangle. So, for \( 36^{\circ}\), we can form 2 triangles? Wait, no, the question is "how many triangles can be made". Wait, maybe I made a mistake. Wait, when two sides are equal (length 5 cm), and we have a given angle:
The formula for the number of triangles:
- If the angle is acute and not the included angle:
- Let the two equal sides be \( a = b\). If \( \angle C=\theta\) (opposite side \( c \)), and \( \theta\) is acute. If \( \theta\) is not the included angle (i.e., \( \theta\) is at \( B \) or \( C \) when \( a = b \)).
Wait, maybe a better approach:
For a triangle with two sides \( a = b = 5\) cm.
- When the angle \( \theta = 36^{\circ}\):
- If \( \theta\) is the included angle (between \( a \) and \( b \)): 1 triangle (SAS).
- If \( \theta\) is an angle opposite one of the equal sides: Since \( a = b \), the angles opposite them are equal. So if \( \theta\) is an angle opposite a 5 - cm side, then the other non - given angle is also \( \theta\), and the third angle is \( 180 - 2\theta\). As long as \( 180-2\theta>0\), we can form a triangle. For \( \theta = 36^{\circ}\), \( 180-2\times36=108^{\circ}>0\), so we can form a triangle. So in total, for \( 36^{\circ}\), we can form 2 triangles? Wait, no, the answer is actually:
Wait, let's recall the triangle construction rules. When two sides are equal (length \( s \)) and we have an angle \( \alpha\):
- If \( \alpha\) is the included angle: 1 triangle (SAS).
- If \( \alpha\) is an angle opposite one of the equal sides:
- If \( \alpha<90^{\circ}\), we can form 2 triangles (one with \( \alpha\) as the base angle and one with \( \alpha\) as the vertex angle? No, wait, no. Wait, when \( \alpha\) is the angle opposite a equal side, since the two sides are equal, the angles opposite them are equal. So if \( \alpha\) is one of those equal angles, then the third angle is \( 180 - 2\alpha\). So we can form one triangle in this case. Wait, I'm getting confused. Let's use the formula for the number of triangles with two equal sides \( a = b \) and a given angle \( \theta \):
- If \( \theta>90^{\circ}\):
- The angle \( \theta\) must be the vertex angle (because if it were a base angle, \( 2\theta>180^{\circ}\), which is not possible). So for \( \theta = 144^{\circ}\), since \( 2\times144=288>180\), the \( 144^{\circ}\) angle must be the vertex angle (between the two equal sides), so we can form 1 triangle.
- If \( \theta = 90^{\circ}\):
- If \( \theta\) is the included angle, we have an isosceles right - triangle (SAS), so 1 triangle. If \( \theta\) is a base angle, \( 2\times90 = 180\), which is not possible (since the sum of angles in a triangle is \( 180^{\circ}\), and we would have a degenerate triangle). So for \( \theta=90^{\circ}\), we can form 1 triangle.
- If \( \theta<90^{\circ}\) (like \( 36^{\circ}\)):
- If \( \theta\) is the included angle: 1 triangle (SAS).
- If \( \theta\) is a base angle: We can form another triangle (since \( 180 - 2\theta>0\)). So total of 2 triangles.

Step2: Analyze 90° case

For \( \theta = 90^{\circ}\):
- If we assume the \( 90^{\circ}\) angle is between the two 5 - cm sides, we can form an isosceles right - triangle (SAS), with legs of length 5 cm.
- If we assume the \( 90^{\circ}\) angle is not between the two 5 - cm sides, then we would have two angles of \( 90^{\circ}\) (since the two sides are equal), and \( 90 + 90=180\), which is impossible (the sum of angles in a triangle is \( 180^{\circ}\), and the third angle would be \( 0^{\circ}\)). So only 1 triangle can be formed.

Step3: Analyze 144° case

For \( \theta = 144^{\circ}\):
- If we assume the \( 144^{\circ}\) angle is not between the two 5 - cm sides, then we would have two angles of \( 144^{\circ}\) (since the two sides are equal), and \( 144+144 = 288>180\), which is impossible.
- So the \( 144^{\circ}\) angle must be between the two 5 - cm sides (the included angle). By SAS, we can form 1 triangle.

Answer:

For \( 36^{\circ}\): 2
For \( 90^{\circ}\): 1
For \( 144^{\circ}\): 1