Chapter 11.1 - Construction of Triangle when two sides and their included angle are given

In the previous section we saw the method of construction of a triangle when it's three sides are given. In this section we will learn another method.

Construction of a Triangle when lengths of 2 sides and included angle are given

Mr. A has now put forward a new challenge. He has a drawing of a triangle XYZ 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 want us to draw an exact replica of the triangle. He will give us one information: The lengths of two sides, and the included angle between those two sides. With this information, can we draw an exact replica? Let us try:

• The given lengths are XY = 8.0 cm, XZ = 6.6 cm and ∠X = 65^o
• We must draw a rough fig. with this given data. This is shown in the fig.11.6. Such a rough fig. will give us an idea about how to proceed

Fig.11.6

• Let us begin the construction. The steps are shown in the fig.11.7
• First we draw a horizontal line 8.0 cm in length, and name it as XY. This is one side of the required triangle. It also fixes two corners X and Y. 
• Now, if we can locate the correct position of ‘Z’, the problem is solved. So our next aim is to locate ‘Z’. The construction steps are shown in the fig.11.7.

Fig.11.7

• From the rough fig., it is clear that, Z lies some where on a line which is inclined at an angle of 65^o to XY. 
• So we draw a line XZ' (of any convenient length) at an angle of 65^o to XY
• With X as center, draw an arc with radius = 6.6 cm
• This arc (shown in green colour) will intersect XZ' at a point. This point of intersection will be at a distance of 6.6 cm from X. So this is our required point Z.
• Join Z to X. This gives us the required ΔXYZ

It may be noted that, this method of constructing a triangle is related to the SAS criterion for congruence, that we learned in the previous chapter. The relation can be explained as follows:
The Δ ABC that we have constructed, and the ΔABC which Mr. A is holding, have two sides and their included angle in common. If two triangles have two sides and their included angle 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 sides, and their included angle, we can easily do the construction.

Solved example 11.3
Construct ΔABC in which AB = 4.8 cm, AC = 6.2 cm, and ∠A = 120^o
Solution:
• First of all we have to draw a rough fig. using the given data. It is shown in the fig. 11.8(a)

Fig.11.8

• Based on the rough fig., we can proceed to do the construction:
• First draw a horizontal line AB of length 4.8 cm
• Draw a line AC' (of any convenient length) at an angle of 120^o to AB
• With A as center, draw an arc of radius 6.2 cm
• This arc (shown in green colour) will intersect AC' at a point. This point of intersection will be at a distance of 6.2 cm from A. So this is our required point C.
• Join B to C. This completes the required ΔABC
Solved example 11.4
ΔABC is an isosceles triangle. The equal sides are AC and BC with lengths of 7.0 cm each. Construct the triangle if ∠B is 42^o
Solution:
• First of all we have to draw a rough fig., using the given data. It is shown in the fig. 11.9(a)

Fig.11.9

• We are given two sides: AC and BC. But their included angle is not given. With out the included angle, we cannot use the two sides to construct the triangle.
• ΔABC is an isosceles triangle. ∠B is one 'base angle'. So the other base angle ∠A will also be same as ∠B. (Details here) So we get ∠A = ∠B = 42^o
• ∠So C = 180 - (42 + 42) = 180 -84 = 96^o (∵ sum of the three interior angles of a triangle is equal to 180^o). Thus the included angle between the two sides = 96^o
• Based on this, we can do the construction. Fig.b shows the steps
• First draw a horizontal line AB' of any convenient length
• At A, draw the line AC' at an angle of 42^o with AB'
• With A as center, draw an arc with a radius of 7.0 cm
• This arc (shown in yellow colour) will cut AC' at a point. This point of intersection is at a distance of 7.0 cm from A. So this point of inter section is C
• With C as center, draw an arc of radius 7.0 cm
• This arc (shown in green colour) will cut the line AB' at B. This point of intersection is at a distance of 7.0 cm from C. So this point is B.
• Join B to C. This completes the required ABC
• An alternate method is to draw a line at an angle of  96^o with AC. This will cut the line AB' at B

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So we have learned the method to construct a triangle when any two of it's sides and their included angle are given. In the next section, we will learn another method.

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Chapter 10.3 - SAS Criterion for the congruence of Triangles

In the previous section we have learned the SSS 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, PQR, STU, UVW and MNO.

Fig.10.12

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. But in this situation, it is not possible because the measurements are incomplete. In such situations, we must try to use other criteria.

We are going to use a rule known as the SAS criterion. The 'A' stands for angle, and, as before, the ‘S’ stands for ‘Side’. So this is the abbreviation for 'Side Angle Side criterion'. Two sides and one angle. 
■ This rule is based on the fact that, when two triangles are congruent, any two sides, and the angle included in between those sides, 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 sides and the included angle: AB = 3.2 cm, AC = 3.9 cm and the included ∠CAB = 72^o
• In fig.(b) we have one particular triangle, which is ΔPQR with the two sides and the included angle: RP = 3.2 cm, RQ = 3.9 cm and the included ∠QRP = 72^o
• The two sides and the included angle are the same. So they are congruent.
■ We are able to establish congruence just by using two sides and the included angle. 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:
• 'SAS' denotes two sides and one angle. This angle should be the included angle between the chosen sides. If we take any other angle, the congruence will not work.
• In this problem, we chose the sides AB and AC, and the ∠CAB between them. We used them for the comparison, and arrived at the conclusion that ΔABC and ΔPQR are in congruence.
• If ΔABC and ΔPQR are in congruence, there will surely be other two SAS combinations also. We will write all the three:
■ [AB, AC, and the included ∠CAB] has a corresponding combination which we already found out: [RP, RQ, and ∠QRP]
■ [CA, CB, and the included ∠ACB] will have a corresponding combination, which we are yet to find
■ [BA, BC, and the included ∠ABC] 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.10.13 below:

Fig.10.13

• In the fig.10.13, the sides 3.9 and 3.2 meet at A and R. So it is obvious that R corresponds to A. That is., A↔R.
• B is reached when we travel 3.2 cm from A. Similarly, P is reached when we travel 3.2 cm from R. So we get B↔P
• C is reached when we travel 3.9 cm from A. Similarly, Q is reached when we travel 3.9 cm from R. So we get C↔Q
• So the correspondence is: A↔R,  B↔P, and  C↔Q. This is same as ABC↔RPQ
• Thus we can write: ΔABC and ΔPQR are congruent to one another under the correspondence ABC↔RPQ
• This is same as writing: ΔABC ≅ ΔRPQ
Now, we use the rough sketch in fig.10.13 above to write the two missing details:
■ [CA, CB, and the included ∠ACB] in ΔABC has a corresponding combination in ΔPQR. What is it?
• We have C↔Q and A↔R. So CA↔QR
• We have C↔Q and B↔P. So CB↔QP
• We have C↔Q. So ∠ACB↔∠RQP
• So the corresponding combination is: [QR, QP and the included ∠RQP]
■ [BA, BC, and the included ∠ABC] in ΔABC has a corresponding combination in ΔPQR. What is it?
• We have B↔P and A↔R. So BA↔PR
• We have B↔P and C↔Q. So BC↔PQ
• We have B↔P. So ∠ABC↔∠RPQ
• So the corresponding combination is: [PR, PQ and the included ∠RPQ]

So we have established the congruence, and put every relations between the two triangles in order. The following animation shows the superposition of ΔRPQ over ΔABC:

Fig.10.14

Based on the above discussion, we can write down the criterion:
■ SAS Congruence criterion:
If under a correspondence, 'two sides and the angle included between them' of a triangle are equal to 'two corresponding sides and the angle included between them' of another triangle, then the triangles are congruent.

Solved example 10.6
(i) In ΔABC, AB = 12 cm, BC = 8 cm, ∠B = 42^o 
In ΔPQR, PQ = 8 cm, QR = 12 cm, ∠Q = 42^o 
Check whether ΔABC and ΔPQR are congruent or not. If they are congruent, write the correspondence.

(ii) In ΔABC, AB = 5 cm, AC = 7 cm, ∠A = 35^o 
In ΔXYZ, XY = 7 cm, XZ = 5 cm, ∠X = 40^o 
Check whether ΔABC and ΔXYZ are congruent or not. If they are congruent, write the correspondence.

(iii) In ΔABC, BC = 14 cm, AC = 8 cm, ∠B = 30^o 
In ΔXYZ, XY = 8 cm, XZ = 14 cm, ∠X = 30^o 
Check whether ΔABC and ΔXYZ are congruent or not. If they are congruent, write the correspondence.


Solution:
(i) We must first draw a rough sketch as shown in fig.10.15:

Fig.10.15

From the rough sketch, it is clear that, 2 sides of 12 cm and 8 cm, and an included angle of 42^o is present in both the triangles. So they are congruent. 
Now we must write the correspondence:
• The two given sides in fig.(a) intersect at B.
• The two given sides in fig.(b) intersect at Q
■ So we get B↔Q
• In fig.(a), if we start from B and move 12 cm, we reach A
• In fig.(b), if we start from Q and move 12 cm, we reach R
■ So we get A↔R
• In fig.(a), if we start from B and move 8 cm, we reach C
• In fig.(b), if we start from Q and move 8 cm, we reach P
■ So we get C↔P
• So the correspondence is: B↔Q, A↔R, and C↔P. This is same as ABC↔RQP
■ Thus we can write: ΔABC and ΔPQR are congruent under the correspondence ABC↔RQP
■ This is same as: ΔABC ≅ ΔRQP

(ii) First we must draw a rough sketch as shown in fig.10.16:

Fig.10.16

From the rough sketch, we can see that 5 cm and 7 cm are present in both the triangles. But the included angles are different. So we cannot say that the two triangles are congruent.

(iii) First we must draw a rough sketch as shown in fig.10.17:

Fig.10.17

From the rough sketch, we can see that 8 cm and 14 cm are present in both the triangles. But the included angle between 8 and 14 for the ΔABC is not given. With out that angle, we cannot check the congruence. So we are not able to say whether the two triangles are congruent or not.
Solved example 10.7
ΔABC in fig.10.18 below is an equilateral triangle. CD and CE divide the ∠C into 3 equal parts. If CD = CE, prove that AD and EB have the same length.

Fig.10.18

Fig.10.19

Solution: 
• Given that ΔABC is an equilateral triangle. So AB = BC = AC
• Given CD and CE divides C into 3 equal parts. So ∠ACD = ∠DCE = ∠ECB
• Given CD = CE
• From the above 3 points, we get 2 equal sides: [AC = BC] and [CD = CE]
• We also get two equal included angles: [∠ACD = ∠BCE]
This is shown in the fig.10.19 above.
• So we have an SAS congruence: ΔADC and ΔBEC are congruent.
• Now we have to write the correspondence:
■ Take the first triangle ADC. In it, AC and DC meet at C
The corresponding sides BC and EC of the second triangle, meet at C.
■ So C↔C
• From C, we move a 'certain distance' to reach A
• From C, we move the same distance to reach B
■ So we get A↔B 
• From C, we move a 'certain distance' to reach D
• From C, we move the same distance to reach E
■ So we get D↔E
• So the correspondence is: C↔C, A↔B, and D↔E. This is same as ACD↔BCE
■ Thus we can write:  ΔACD ≅ ΔBCE
• Now we have to prove that AD = BE:
• From ΔACD, take the corner A. The corresponding point in ΔBCE is B
• From ΔACD, take the corner D. The corresponding point in ΔBCE is E
■ So the side corresponding to AD is BE
• Since the two triangles are congruent, the corresponding sides will be equal
■ Thus we get AD = BE
Solved example 10.8
In the fig.10.20 given below, UV and WX bisect each other at 'O'. 
(i) Prove that ΔOUW and ΔOVX are congruent to one another. 
(ii) Write all the 6 corresponding parts of the two triangles. 
(iii) Pick out the true statements from: (a)  ΔOUW ≅ ΔOXV  (b) ΔOWU ≅ ΔOXV  (c) ΔUOW ≅ ΔVOX

Fig.10.20

Solution:
(i) • UV and WX bisect each other at 'O'. So OU = OV, and OW = OX
• ∠WOU = ∠VOX (∵ they are 'opposite angles')
• We have 'two sides and the included angle' same on the two triangles ΔOUW and ΔOVX, and so, they are congruent.
(ii) Now we have to write the correspondence:
• Take the first ΔOUW: OW and OU meet at 'O'. The corresponding sides in the second triangle ΔOVX are OX and OV respectively. They meet at 'O'
• So we get O↔O
• From O, we move a 'certain distance' to reach U
• From O, we move the same distance to reach V
■ So we get V↔U 
• From O, we move a 'certain distance' to reach W
• From O, we move the same distance to reach X
■ So we get W↔X
• So the corresponding corners are: O↔O, U↔V, W↔X (Angles at these corresponding corners will be equal) - - - (1)
• From this, we can write the corresponding sides: OU↔OV, OW↔OX, UW↔VX (Lengths of these corresponding sides will be equal)
■ Now we can write the congruence statement:
ΔOUW and ΔOVX are congruent under the correspondence: OUW↔OVX
This is same as: ΔOUW ≅ ΔOVX
(iii) (a) ΔOUW ≅ ΔOXV
This is not true because:  O↔O and U↔X are true but W↔V is not true [from (1)]
(b) ΔOWU ≅ ΔOXV
This is true because:  O↔O, W↔X, and U↔V are all true
(c) ΔUOW ≅ ΔVOX
This is true because:  U↔V, O↔O, and W↔X are all true

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So we have completed the discussion on SAS criterion. In the next section we will discuss about another criterion.

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