In the previous section we saw the method of construction of a triangle when any two sides and their included angle are given. In this section we will learn another method.
• The given angles are P = 55o, Q = 40o and PQ = 8.2 cm
• We must draw a rough fig. with this given data. This is shown in the fig.11.10. Such a rough fig. will give us an idea about how to proceed
• Let us begin the construction. The steps are shown in the fig.11.11
• First we draw a horizontal line 8.2 cm in length, and name it as PQ. This is one side of the required triangle. It also fixes two corners P and Q.
• Now, if we can locate the correct position of ‘R’, the problem is solved. So our next aim is to locate ‘R’.
• From the rough fig., it is clear that, R lies some where on a line which is inclined at an angle of 55o to PQ.
• So we draw a line PP' (of any convenient length) at an angle of 55o to PQ
• From the rough fig., it is also clear that, R lies some where on a line which is inclined at an angle of 40o to PQ.
• So we draw a line QQ' (of any convenient length) at an angle of 40o to PQ
• The two lines PP' and QQ' intersect at the point R. So ΔPQR is the required triangle
It may be noted that, this method of constructing a triangle is related to the ASA criterion for congruence, that we learned in the previous chapter. The relation can be explained as follows:
The ΔPQR that we have constructed, and the ΔPQR which Mr. A is holding, have two angles and their included side in common. If two triangles have two angles and their included side the same, they are congruent to one another. In other words, one is the exact replica of the other. So next time some one gives us any two angles, and their included side, we can easily do the construction.
Solved example 11.5
Construct ΔABC in which ∠A = 37o, ∠B = 78o and AB = 6.4 cm
Solution:
• First of all we have to draw a rough sketch using the given data. It is shown in the fig. 11.12(a)
• Based on the rough sketch, we can proceed to do the construction:
• First draw a horizontal line AB of length 6.4 cm
• Draw a line AA' (of any convenient length) at an angle of 37o to AB
• Draw a line BB' (of any convenient length) at an angle of 78o to AB
• These two lines will intersect at C. ΔABC is our required triangle
Solved example 11.6
In ΔMNO, MN = 4.4 cm, ∠N = 101o and ∠O = 48o. Construct the triangle.
Solution:
• First of all we have to draw a rough fig., using the given data. It is shown in the fig.11.13(a)
• We are given one side: MN = 4.4 cm.
• ∠N = 101o. ∠M is not given. We can use the 'ASA congruence' only if we get ∠M
• But ∠O is given as 48o. So M = 180 - (101 + 48) [∵ sum of the three interior angles of a triangle = 180o]
• So we get ∠M = 180 - 149 = 31o
• Now we can use the ASA property:
• First draw a horizontal line MN of length 4.4 cm
• Draw a line MM' (of any convenient length) at an angle of 31o to MN
• Draw a line NN' (of any convenient length) at an angle of 101o to MN
• These two lines will intersect at O. ΔMNO is our required triangle
So we have learned the method to construct a triangle when any two of it's angles and their included side are given. In the next section, we will learn one more method.
Construction of a Triangle when two angles and the included side are given
Mr. A now puts forward another challenge. He has a drawing of a triangle PQR with him. As before, all the details (lengths of all 3 sides, and angles at all 3 corners) of the triangle is given in the drawing. He does not want to show us that drawing. But he wants us to draw an exact replica of the triangle. He will give us one information: 'Two angles', and the 'length of the side included in between those two angles'. With this information, can we draw an exact replica? Let us try:• The given angles are P = 55o, Q = 40o and PQ = 8.2 cm
• We must draw a rough fig. with this given data. This is shown in the fig.11.10. Such a rough fig. will give us an idea about how to proceed
 |
| Fig.11.10 |
• First we draw a horizontal line 8.2 cm in length, and name it as PQ. This is one side of the required triangle. It also fixes two corners P and Q.
• Now, if we can locate the correct position of ‘R’, the problem is solved. So our next aim is to locate ‘R’.
 |
| Fig.11.11 |
• So we draw a line PP' (of any convenient length) at an angle of 55o to PQ
• From the rough fig., it is also clear that, R lies some where on a line which is inclined at an angle of 40o to PQ.
• So we draw a line QQ' (of any convenient length) at an angle of 40o to PQ
• The two lines PP' and QQ' intersect at the point R. So ΔPQR is the required triangle
It may be noted that, this method of constructing a triangle is related to the ASA criterion for congruence, that we learned in the previous chapter. The relation can be explained as follows:
The ΔPQR that we have constructed, and the ΔPQR which Mr. A is holding, have two angles and their included side in common. If two triangles have two angles and their included side the same, they are congruent to one another. In other words, one is the exact replica of the other. So next time some one gives us any two angles, and their included side, we can easily do the construction.
Solved example 11.5
Construct ΔABC in which ∠A = 37o, ∠B = 78o and AB = 6.4 cm
Solution:
• First of all we have to draw a rough sketch using the given data. It is shown in the fig. 11.12(a)
 |
| Fig.11.12 |
• First draw a horizontal line AB of length 6.4 cm
• Draw a line AA' (of any convenient length) at an angle of 37o to AB
• Draw a line BB' (of any convenient length) at an angle of 78o to AB
• These two lines will intersect at C. ΔABC is our required triangle
Solved example 11.6
In ΔMNO, MN = 4.4 cm, ∠N = 101o and ∠O = 48o. Construct the triangle.
Solution:
• First of all we have to draw a rough fig., using the given data. It is shown in the fig.11.13(a)
 |
| Fig.11.13 |
• ∠N = 101o. ∠M is not given. We can use the 'ASA congruence' only if we get ∠M
• But ∠O is given as 48o. So M = 180 - (101 + 48) [∵ sum of the three interior angles of a triangle = 180o]
• So we get ∠M = 180 - 149 = 31o
• Now we can use the ASA property:
• First draw a horizontal line MN of length 4.4 cm
• Draw a line MM' (of any convenient length) at an angle of 31o to MN
• Draw a line NN' (of any convenient length) at an angle of 101o to MN
• These two lines will intersect at O. ΔMNO is our required triangle
So we have learned the method to construct a triangle when any two of it's angles and their included side are given. In the next section, we will learn one more method.
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