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.
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
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| Fig.11.10 |
• 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’.
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| Fig.11.11 |
• 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)
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| Fig.11.12 |
• 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)
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| Fig.11.13 |
• 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.
In the previous section we have learned the SAS Criterion for the congruence of Triangles. In this section we will discuss about another criterion.
Fig.10.12(a) below, shows a triangle ABC. Fig.(b) shows five triangles: XYZ, WYZ, PQR, DEF, and MNO.
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| Fig.10.21 |
Our problem is this: Is there any triangle in fig.(b), which is congruent to ΔABC? If yes, which one?
Solution: Looking at figs.(a) & (b), we find that the measurements of the triangles are incomplete. None of the triangles have all lengths of all sides, or angles at all corners, marked on them.
• If the triangles had the lengths of all the 3 sides, then we could straight away use the SSS criterion to check the congruence.
• If two sides and their included angle is given, we could use the SAS criterion.
But in this situation, these are not possible because the measurements are incomplete. We cannot find an SSS comparison or SAS comparison. In such situations, we must try to use other criteria.
We are going to use a rule known as the ASA criterion. As before, the 'A' stands for angle, and, the ‘S’ stands for ‘Side’. So this is the abbreviation for 'Angle Side Angle criterion'. Two angles and one side.
■ This rule is based on the fact that, when two triangles are congruent, any two angles, and the side between those angles, in one triangle will be present in the other triangle. This rule will become clear when we apply it to our problem:
• In fig.(a), We have ΔABC with two angles and the included side: ∠A = 65o, ∠C = 82o and the included side AC = 2.7 cm
• In fig.(b) we have one particular triangle, which is ΔDEF with the two angles and the included side: ∠E = 65o, ∠D = 82o and the included side ED = 2.7 cm
• The two angles and the included side are the same. So ΔABC and ΔDEF are congruent.
■ We are able to establish congruence just by using two angles and the included side. All other sides and angles may or may not be given in the problem. But we do not need them to establish congruence.
Some important points to note:
• 'ASA' denotes two angles and one side. This side should be the included side between the chosen angles. If we take any other side, the congruence will not work.
• In this problem, we chose the angles ∠A and ∠C, and the side AC between them. We used them for the comparison, and arrived at the conclusion that ΔABC and ΔDEF are in congruence.
• If ΔABC and ΔDEF are in congruence, there will surely be other two ASA combinations also. We will write all the three:
■ [∠A, ∠C, and the included side AC] has a corresponding combination which we already found out: [∠E, ∠D, and the included side ED]
■ [∠B, ∠C, and the included side BC] will have a corresponding combination, which we are yet to find
■ [∠A, ∠B, and the included side AB] will have a corresponding combination, which we are yet to find
So there are two details that we are yet to find. Those two details are not necessary to establish a congruence. But establishing a congruence does not solve the problem completely. We have a little more work to do. We have to put the sides and corners of the two triangles in order. We do this by writing the correspondence. And, after writing the correspondence, we will be able to write those two details very easily.
So let us try to write the correspondence:
The one detail which we already know, can be shown by a rough sketch as in the fig.10.22 below:
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| Fig.10.22 |
• From the fig.10.22, it is obvious that A↔E and C↔D. Because, ∠A and ∠E have the same measure, and similarly, ∠C and ∠D have the same measure
• The only remaining corner in the first triangle is B, and that in the second triangle is F. So we get B↔F
• So the correspondence is: A↔E, B↔F, and C↔D. This is same as ABC↔EFD
• Thus we can write: ΔABC and ΔDEF are congruent to one another under the correspondence ABC↔EFD
• This is same as writing: ΔABC ≅ ΔEFD
Now, we use the rough sketch in fig.10.13 above to write the two missing details:
■ [∠B, ∠C, and the included side BC] in ΔABC has a corresponding combination in ΔDEF. What is it?
• We have B↔F, and C↔D. So BC↔FD
• Thus the corresponding combination is: [∠F, ∠D, and the included side FD]
■ [∠A, ∠B, and the included side AB] in ΔABC has a corresponding combination in ΔDEF. What is it?
• We have A↔E, and B↔F. So AB↔EF
• Thus the corresponding combination is: [∠E, ∠F, and the included side EF]
So we have established the congruence, and put every relations between the two triangles in order. The following animation shows the superposition of ΔDEF over ΔABC:
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| Fig.10.23 |
Based on the above discussion, we can write down the criterion:
■ SAS Congruence criterion:
If under a correspondence, 'two angles and the included side' of a triangle are equal to 'two corresponding angles and the included side' of another triangle, then the triangles are congruent.
Solved example 10.9
There are two triangles. ΔPQR and ΔXYZ. We are required to establish a congruence: ΔPQR ≅ ΔYXZ. It is known that PQ = YX. What additional information is required to establish the congruence?
Solution:
• The required congruence is: ΔPQR ≅ ΔYXZ
• So P↔Y, Q↔X and R↔Z
• It is known that PQ = YX
• Let us draw a rough sketch as shown below:
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| Fig.10.24 |
• From the fig., it is clear that, if we are given an additional information that, ∠P = ∠Y, and ∠Q = ∠X, the two triangles will be congruent (ASA congruence)
Solved example 10.10
In fig.10.25, UW = VX. Establish the congruence between ΔOUW and ΔOVX, and write the correspondence
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| Fig.10.25 |
Solution:
• In the two triangles, UW = VX. So we have one side equal.
• If we can prove that the 'corresponding angles at the ends of these sides' also equal, then the triangles will be congruent by the ASA rule. So let us try to prove it:
• We have ∠U + ∠O = 27 + 37 = 64o
• ∴ ∠W = 180 - 64 = 116o (∵ sum of the interior angles of a triangle is 180o)
• ∠UOW = ∠VOX = 37o (∵ they are opposite angles )
• In ΔOVX, ∠O + ∠V = 37 + 27 = 64o
• ∴ ∠X = 180 - 64 = 116o (∵ sum of the interior angles of a triangle is 180o)
• Thus we get ∠W = ∠X. We already have ∠U = ∠V and lengths of sides UW = XV
• So ΔOUW and ΔOXV are congruent according to the ASA congruence criterion
• Now we have to write the correspondence:
• We have ∠W = ∠X and ∠U = ∠V
• So W↔X and U↔V. The only remaining corner is O. So O↔O
• This can be written as OUW↔OVX
• So we can write: ΔOUW and ΔOXV are congruent under the correspondence OUW↔OVX
• This can also be written as ΔOUW ≅ ΔOVX
Solved example 10.11
Given below are some pairs of triangles. In each pair, examine if the triangles are congruent to one another. If they are congruent, write the correspondence.
(i) ΔABC: ∠A = 30o, ∠B = 45o, AB = 4 cm. │ΔXYZ: ∠Y = 30o, ∠Z = 45o, YZ = 4 cm
(ii) ΔPQR: ∠P = 40o, ∠Q = 60o, PQ = 7 cm. │ΔABC: ∠A = 40o, ∠B = 60o, AC = 7 cm
(iii) ΔXYZ: ∠X = 80o, ∠Y = 50o, XY = 6 cm. │ΔPQR: ∠P = 80o, ∠R = 50o, PQ = 6 cm
Solution:
(i) A rough sketch is shown in the fig.10.26(i)
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| Fig.10.26 |
• Side AB = YZ, ∠A = ∠Y and ∠B = ∠Z. So the triangles are congruent by ASA criterion
• Now we write the correspondence:
• ∠A = ∠Y = 30o. So A↔Y.
• ∠B = ∠Z = 45o. So B↔Z
• The remaining corners are C and X. So C↔X.
• Thus we have A↔Y, B↔Z, C↔X ⇒ ABC↔YZX
• So ΔABC and ΔXYZ are congruent under the correspondence ABC↔YZX
• This is same as ΔABC ≅ ΔYZX
(ii) A rough sketch is shown in the fig.10.26(ii)
• ∠P = ∠A and ∠Q = ∠B. But length of AB is not given.
• AC is given as 7 cm. If ∠C = 60o, AC will correspond to PQ
• So we have to find ∠C:
• ∠C = 180 - (40 + 60) (∵ sum of the interior angles of a triangle is 180o)
• So ∠C = 180 - 100 = 80o
• So ∠C is not 60o, and thus, without the length of AB, we can not say if the two triangles are congruent or not
(iii) A rough sketch is shown in the fig.10.26(iii)
• ∠X = ∠P and ∠Y = ∠R. But length of PR is not given.
• PQ is given as 7 cm. If ∠Q = 50o, PQ will correspond to XY
• So we have to find ∠Q:
• ∠Q = 180 - (80 + 50) (∵ sum of the interior angles of a triangle is 180o)
• So ∠Q = 180 - 130 = 50o
Q is indeed equal to 50. So we have an ASA congruence
• Now we write the correspondence:
• ∠X = ∠P = 80o. So X↔P.
• ∠Z = ∠Q = 50o. So Z↔Q
• The remaining corners are Y and R. So Y↔R.
• Thus we have X↔P, Y↔R, Z↔Q ⇒ XYZ↔PRQ
• So ΔXYZ and ΔPQR are congruent under the correspondence XYZ↔PRQ
• This is same as ΔXYZ ≅ ΔPRQ
Solved example 10.12
In the fig.10.27 below, the ray OS bisects ∠POR. It also bisects ∠PQR. Prove that OP = OR and PQ = RQ
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| Fig.10.27 |
Solution:
• The ray OS bisects ∠ROP. So we get ∠ROQ = ∠POQ
• The ray OS bisects ∠PQR. So we get ∠PQO = ∠RQO
• OQ is common. So we have an ASA congruence
• The triangles ΔOQR and ΔOQP are congruent.
• Now we have to write the correspondence:
■ For the ASA congruence, we took [∠ROQ, ∠OQR, and their included side OQ] in ΔOQR
■ We took [∠POQ, ∠OQP and their included side OQ] in ΔOQP
• The above two are the corresponding combinations
• In those corresponding combinations, ∠ROQ = ∠POQ. So we get O↔O
• Also ∠OQR = ∠OQP. So we get Q↔Q
• The only remaining corners are P and R So we get P↔R
• Thus we can write OPQ↔ORQ
• So the two triangles ΔOPQ and ΔOQR are congruent under the correspondence: OPQ↔ORQ
• This is same as ΔOPQ ≅ ΔORQ
• Now we have to prove that OP = OR
• We have already proved that O↔O and P↔R
• So OP and OR are corresponding sides. Corresponding sides of congruent triangles will be equal in length. Thus OP = OR
• Next we have to prove that PQ = RQ
• We have already proved that P↔R and Q↔Q
So PQ and RQ are corresponding sides. Corresponding sides of congruent triangles will be equal in length. Thus PQ = RQ
So we have completed the discussion on ASA criterion. In the next section we will discuss about one more criterion.
