Are you feeling difficulty constructing triangles? If yes, then need not worry about it. Constructions are the easiest and scoring topic in maths. Learn how to Construct a Triangle when Two of its Angles and the included Side are given from here. See the Question and Answers on constructing a triangle when two of its angles and the included Side are given.

Two angles and side is known as ASA (Angle-Side-Angle). See the construction steps from this page. We will help you to construct a triangle with two angles and one side in a simple method. Get the step-by-step explanation for all the questions here.

**Do Refer:**

- To Construct a Triangle whose Three Sides are given
- To Construct a Triangle when Two of its Sides and the included Angles are given

## How to Construct a Triangle when Two Angles and One Side(ASA) is given?

There are four steps to construct a triangle.

Step 1. Draw the line segment.

Step 2. Using a protractor draw a ray making an angle with the line segment.

Step 3. Using a protractor at another point draw another arc.

Step 4. By using the property “Sum of the three angles of any triangle is 180Â°. With this property, we can find the third angle.

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### Construction of Triangle with Two Angles and One Side Examples

**Example 1.**

Draw a triangle ABC in which âˆ B = 60Â° and âˆ C = 70Â° and the side BC = 6 cm.

**Solution:**

Step 1. Draw the line segment BC.

Step 2. Using a protractor draw a ray making a 60Â° angle with the line segment BC = 6 cm.

Step 3. Using a protractor draw another angle with a measure of 70 degrees at another point draw another arc with C as the center.

Step 4. By using the property “Sum of the three angles of any triangle is 180Â°. With this property, we can find the third angle.

Step 5. Join PB and QC.

Thus the required triangle ABC is formed.

**Example 2.**

Draw a triangle ABC in which âˆ B = 40Â° and âˆ C = 50Â° and the side BC = 8 cm.

**Solution:**

Step 1. Draw the line segment BC = 8cm.

Step 2. Using a protractor draw a ray making a 40Â° angle with the line segment BC = 8 cm.

Step 3. Using a protractor draw another angle with a measure of 70 degrees at another point draw another arc with C as the center.

Step 4. By using the property “Sum of the three angles of any triangle is 180Â°. With this property, we can find the third angle.

Step 5. Join PC and QB.

Thus the required triangle ABC is formed.

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**Example 3.**

Draw a triangle XYZ in which âˆ Y = 100Â° and âˆ X = 50Â° and the side XY = 8 cm.

**Solution:**

Step 1. Draw the line segment XY = 5cm.

Step 2. Using a protractor draw a ray making a 100 Â° angle with the line segment XY = 8 cm.

Step 3. Using a protractor draw another angle with a measure of 50 degrees at another point draw another arc with Y as the center.

Step 4. By using the property “Sum of the three angles of any triangle is 180Â°. With this property, we can find the third angle.

Step 5. Join PX and QY.

Thus the required triangle XYZ is formed.

**Example 4.**

Draw a triangle ABC in which âˆ CAB = 55Â° and âˆ CBA = 70Â° and the side AB = 5 cm.

**Solution:**

Step 1. Draw the line segment AB = 5cm.

Step 2. Using a protractor draw a ray making a 55Â° angle with the line segment AB = 5 cm.

Step 3. Using a protractor draw another angle with a measure of 70 degrees at another point draw another arc with B as the center.

Step 4. By using the property “Sum of the three angles of any triangle is 180Â°. With this property, we can find the third angle.

Step 5. Join AQ and PB.

Thus the required triangle ABC is formed.

**Example 5.**

Draw a triangle PQR in which âˆ RPQ = 80Â° and âˆ RQP = 55Â° and the side PQ = 8 cm

**Solution:**

Step 1. Draw the line segment PQ = 8cm.

Step 2. Using a protractor draw a ray making a 80Â° angle with the line segment PQ = 8 cm.

Step 3. Using a protractor draw another angle with a measure of 55 degrees at another point draw another arc with Q as the center.

Step 4. By using the property “Sum of the three angles of any triangle is 180Â°. With this property, we can find the third angle.

Step 5. Join PR and RQ.

Thus the required triangle PQR is formed.