A plane figure bounded by three non concurrent line segment in plane is called a TRIANGLE.
Clearly, when we say a plane figure, we actually mean the linear figure, not the two dimensional figure. Let A, B, C are non linear points such that they are not on the same line; we say A, B, C are noncollinear. Join AB, BC , CA. We get a linear figure which consists of three line segments which meet only at their end points. Such a linear figure is called a triangle. We say A, B and C are the vertices of the triangle ABC. The segments AB, BC and CA are called the sides of the triangle ABC; and the angles ∠ABC, ∠BCA, ∠CAB are the three angles of a triangle ABC.
A triangle consists of three elements:
Vertices | Sides | Angles |
A | AB | ∠BAC or ∠A |
B | BC | ∠ABC or ∠B |
C | CA | ∠BCA or ∠C |
Triangles are classified based on the measure of sides and angles of the triangle.
Classification based on sides of the triangle:
There are three types:
(i) Equilateral triangle
(ii) Isosceles triangle
(iii) Scalene triangle
Now let us study in brief:
Classification based on angles of the triangle:
There 3 types of triangle:
(i) Acute angled triangle
(ii) Right angled triangle
(iii) Obtuse angled triangle
Exercise 3.2.1
1. Match the following:
(i)
(a) Equilateral triangle
(ii)
(b) Acute angled triangle
(iii)
(c) Right angled triangle
(iv)
(d) Obtuse angled triangle
Solution:
————— Right angled triangle
—————– Obtuse angled triangle
——————- equilateral triangle
——————— Acute angled triangle
2. Based the sides, classify the following triangles:
(i)
Solution:
Scalene triangle [all the three sides are different from one another]
(ii)
Solution:
Scalene triangle [ all the three sides are different from one another]
(iii)
Solution:
Scalene triangle [ all the three sides are different from one another]
(iv)
Solution:
Isosceles triangle [ the two sides of triangle are equal to one another]
(v)
Solution:
Scalene triangle [ all the three sides are different from one another]
(vi)
Solution:
Scalene triangle [ all the three sides are different from one another]
(vii)
Solution:
Scalene triangle [ all the three sides are different from one another]
(viii)
Solution:
Equilateral triangle [ all the three sides are equal to one another]
(ix)
Solution:
Isosceles triangle [ the two sides of triangle are equal to one another]
(x)
Solution:
Isosceles triangle [ the two sides are triangle are equal to one another]
3.2.2 Sum of interior angles
Theorem 1: In any triangle, the sum of the three interior angles is 180°
This theorem is also called interior angle theorem.
Given: ABC is a triangle.
To prove: ∠ABC + ∠CAB + ∠ACB = 180°
Construction: through point A draw the line EF||BC
Proof:
Below we give several statements and the reason for the truth of each statement. Finally we arrive at the desired conclusion.
Statement:
∠ABC = ∠EAB [alternate angles by the transversal AB with the parallel lines BC and EF]
∠BCA= ∠FAC [alternate angles by the transversal AB with the parallel lines BC and EF]
∠EAB + ∠BAC + ∠FAC = 180 ° [Sum of all linear angles at A]
By substituting ∠EAB = ∠ABC and ∠FAC = ∠BCA, we finally get,
∠ABC + ∠CAB + ∠ACB = 180°
This completes the proof.
Example 1: In a triangle ABC, it is given that ∠B = 105° and ∠C = 50°. Find ∠A.
Solution:
We have, in triangle ABC, [by theorem 1]
∠A + ∠B + ∠C = 180°
We know,
∠B = 105° and ∠C = 50°
Therefore, ∠A = 180° – ∠B – ∠C
∠A = 180° – 105° – 50°
∠A = 25°
Thus, ∠A Measures 25°
Example 2: In the figure, find all the angles:
Solution:
In triangle ABC, if we make use of theorem 1 [interior angle theorem], we get,
∠A + ∠B + ∠C = 180° ——————– (1)
In the figure it is given that, ∠A = 5x , ∠B = 3x, ∠C = 2x
Let us substitute the value of ∠A, ∠B and ∠C in equation (1)
Then, we get,
5x + 3x + 2x = 180°
10x = 180°
Therefore, x = 18°
Thus, ∠A = 5x = 5 x 18° = 90°
∠B = 3x = 3 x 18° = 54°
∠C = 2x = 2 x 18° = 36°
Example 3: If the bisectors of the angles ∠ABC and ∠ACB of a triangle meet at a point O, then prove that ∠BOC = 90° + ½ ∠BAC
Solution:
Given: A triangle ABC and the bisectors of ∠ABC and ∠ACB meeting at point O.
To prove: ∠BOC = 90° + ½ ∠BAC
Proof: In triangle BOC, we have,
∠1 + ∠2 + ∠BOC = 180° ———-(1)
In triangle ABC, we have, ∠ABC + ∠CAB + ∠ACB = 180°
Since, BO and CO are bisectors of ∠ABC and ∠ACB respectively, we have,
∠B = 2∠1 and ∠C = 2∠2
Therefore, we get, ∠A + 2∠1 + 2∠2 = 180°
Let us divide it by 2, we get,
∠1 + ∠2 = 90° – ∠A /2 ——————-(2)
From (1) and (2), we get,
90° – ∠A /2 + ∠BOC = 180°
Hence, ∠BOC = 90° + ½ ∠BAC
Exercise 3.2.2
- In a triangle ABC, if ∠A = 55° and ∠B = 40°, find ∠C
Solution:
We know, ∠A + ∠B + ∠C = 180° ———– (1) [from theorem 1 – interior angle theorem]
It is given that,
∠A = 55° and ∠B = 40°
Substitute the value of ∠A and ∠B in equation (1)
Then,
55° + 40° + ∠C = 180°
∠C = 180° – 55° – 40°
∠C = 85°
- In a right angled triangle, if one of the other two angles is 35° , find the remaining angle:
Solution:
Data: triangle ABC is a right angled triangle and ∠C =35°
Now, we have find ∠C
We know, ∠A + ∠B + ∠C = 180° —————–(1)
Let us substitute the value of ∠A and ∠B in equation (1)
∠C = 180° – 90° – 35°
∠C = 55°
Thus, ∠C measures 55°
3. If the vertex angle of an isosceles triangle is 50°, find the other angles.
Solution:
Data: Triangle ABC is an isosceles triangle, AB = AC and ∠A = 50°
We know, AB = AC, therefore ∠B = ∠C
We know, ∠A + ∠B + ∠C = 180° ———– (1) [from theorem 1 – interior angle theorem]
∠A = 50°
Substitute the value of ∠A in equation (1)
Then,
50° + ∠B + ∠C = 180°
We know, ∠B = ∠C
∠B + ∠B = 180° – 50°
2∠B = 130°
∠B = 130°/2 = 65°
∠B = ∠C = 65°
4. The angles of a triangle are in the ratio 1:2:3. Determine the three angles.
Solution:
Let ∠A = 1x, ∠B = 2x, ∠C = 3x
Then,
∠A + ∠B + ∠C = 180° ———– (1) [from theorem 1 – interior angle theorem]
Substitute the value of ∠A, ∠B and ∠C in equation (1)
Then,
x + 2x + 3x = 180°
6x = 180°
x = 180°/6
x = 30°
Therefore,
∠A = 30°,
∠B = 2x = 2 x 30 = 60°,
∠C = 3x = 3 x 30 = 90°
5. In adjacent triangle ABC, find the value of x and calculate the measure of all the angles of the triangle.
Let ∠A = x + 15, ∠B = x – 15, ∠C = x+30
Then,
∠A + ∠B + ∠C = 180° ———– (1) [from theorem 1 – interior angle theorem]
Substitute the value of ∠A, ∠B and ∠C in equation (1)
Then,
(x + 15) + (x -15) + (x + 30) = 180°
3x + 30 = 180°
3x = 150
x = 150°/3
x = 50°
Therefore,
∠A = x + 15° = 50° + 15° = 65°,
∠B = x – 15° = 50° – 15° = 35°,
∠C = x + 30° = 50° + 30° = 80°
6. The angles of a triangle are arranged in ascending order of their magnitude. If the difference between two consecutive angles is 10°, find the three angles.
Solution:
Let ∠A = x + 10°, ∠B = x + 20°, ∠C = x + 30°
Then,
∠A + ∠B + ∠C = 180° ———– (1) [from theorem 1 – interior angle theorem]
Substitute the value of ∠A, ∠B and ∠C in equation (1)
Then,
(x + 10°) + (x + 20°) + (x + 30°) = 180°
3x + 60 = 180°
3x = 120°
x = 120°/3
x = 40°
Therefore,
∠A = x + 10° = 40° + 10° = 50°,
∠B = x + 20° = 40° + 20° = 60°,
∠C = x + 30° = 40° + 30° = 70°
3.2.3 Exterior angles
Consider a triangle ABC. If the side BC is produced externally to form a ray BD, then ∠ACD is called an exterior angle of triangle ABC at C and is denoted by Ext∠C
Theorem 2: If a side of triangle is produced, the exterior angle so formed is eequal to the sum of corresponding interior opposite angles. [ Exterior angle theorem]
Given: In triangle PQR, produce QR to S. Then ∠PRS is an exterior opposite angles are ∠PRS and ∠QPR.
To prove: ∠PRS = ∠QPR + ∠PQR
Proof:
Statement:
∠QPR + ∠PQR + ∠PRQ = 180° [Interior angle theorem]
∠PRQ + ∠PRS = 180° [linear pair]
∠QPR + ∠PQR + ∠PRQ = ∠PRQ + ∠PRS [Axiom1 – Things which are equal to the same thing are equal to one another.]
∠QPR + ∠PQR = ∠PRS [Axiom 2 – If equals are added to equals, the wholes are equal.]
This completes the proof.
Example 4: An exterior angle of a triangle is 100° and one of the interior opposite angles is 45°. Find the other two angles of the triangle.
Solution:
Let ABC be a triangle whose side BC is produced to form an exterior angle ∠ACD such that, ext∠ACD is 100°. Let ∠B = 45°.
By exterior angle theorem we have, ∠ACD = ∠B + ∠A
100° = 45° + ∠A
∠A = 100° – 45° = 55°
Hence, ∠A = 55°
∠C = 180° – (∠A + ∠B)
= 180° – (55° + 45°)
= 80°
Example 5: In the given figure, sides QR and RQ of a triangle PQR are produced to the points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ
Solution:
Since Q, P and S all lie in the same plane,
∠QPR + ∠SPR = 180°
Hence, ∠QPR + 135° = 180°
or
∠QPR = 180° – 135° = 45°
Usinf exterior angle theorem in triangle PQR, we have
∠PQT = ∠QPR + ∠PRQ
This gives, 110° = 45° + ∠PRQ
Solving for, ∠PRQ, we get, ∠PRQ = 110° – 45° = 65°
Example 6: The side BC of a triangle ABC is produced on both the sides. Show that the sum of the exterior angles so formed is greater than ∠A by two right angles.
Solution:
Let us draw a triangle ABC, and produce BC on both sides to points D and F. Denote the angles as shown in the figure.
Now, we have to show that, ∠4 + ∠5 = ∠1 + 180°
By exterior theorem, we have,
∠4 = ∠1 + ∠3 and ∠5 = ∠1 + ∠2
Adding these two , we get,
∠4 + ∠5 = (∠1 +∠3) + (∠1 + ∠2) = ∠1 +(∠1 + ∠2 + ∠3) = ∠1 + 180°
Since the sum of all the interior angles of a triangle is 180°.
Exercise 3.2.3
- The exterior angles obtained on producing the base of a triangle both ways are 104° and 136°. Find the angles of the triangle.
Solution:
Let ABC be a given triangle with base BC. B is produced till E and C is produced till F this makes ∠ABF = 104° and ∠ACF = 136°. Now we have to find ∠ABC, ∠BCA and ∠ACB.
∠ACB + ∠ACF = 180° (linear pair)
We know,
∠ACF = 136°
Then, ∠ACB = 180° – ∠ACF = 180° – 136° = 44°
∠ABE + ∠ABC = 180° (linear pair)
We know,
∠ABE = 104°
Then, ∠ABC = 180° – ∠ABE = 180° – 104° = 76°
From interior angles theorem , sum of three interior angles is equal to 180°
Then,
∠ABC + ∠BCA + ∠CAB = 180° —————(1)
We know,
∠ABC = 76° and ∠ACB = 44°
Substitute the value of ∠ACB = 44° and ∠ABC = 76° in equation (1)
76° + 44° + ∠CAB = 180°
Then, ∠CAB = 180° – 76° – 44° = 60°
Therefore, ∠ABC = 76°
∠BCA = 44°
∠CAB = 60°
2. Sides BC, CA and AB of a triangle ABC are produced in an order, forming exterior angles ∠ACD, ∠DAE and ∠CBF. Show that ∠ACD + ∠DAE + ∠CBF = 360˚.
Solution:
∠ACB + ∠ACD = 180˚ ( Proposition 1: Let AB be a straight line and OC be a ray standing on the line AB. Then ∠BOC + ∠COA = 180°)
∠BAC + ∠CAE = 180˚ (Proposition 1: Let AB be a straight line and OC be a ray standing on the line AB. Then ∠BOC + ∠COA = 180°)
∠FBC + ∠ABC = 180˚ (Proposition 1: Let AB be a straight line and OC be a ray standing on the line AB. Then ∠BOC + ∠COA = 180°)
Adding above equations.
∠ACB + ∠ACD + ∠BAC + ∠CAE + ∠FBC + ∠ABC = 360˚ + 160˚
∠ACD + ∠CAE + ∠CBF + { ∠ACB + ∠BAC + ∠ABC } = 360˚ + 180˚
∠ACD + ∠BAE + ∠CBF + 180˚ ( Interior angle theorem ) = 360˚ + 180˚
∠ACD + ∠BAE + ∠CBF = 360˚ ( Axiom 3: If equals are subtracted from equals, then the remainders are equal.)
3. Compute the value of x in each of the following figures:
(i))
Solution:
Triangle ABC is an isosceles triangle. AB = AC , then ∠ABC = ∠ACB = 50°
∠ACD = x
∠ACD + ∠ACB = 180°
x = 180° – ∠ACB = 180° – 50° = 130°
(ii)
Solution:
∠ACB = x
∠CAD + ∠CAB = 180°
∠CAB = 180° – ∠CAD = 180° – 130° = 50°
∠ABE + ∠ABC = 180°
∠ABC = 180° – ∠ABE = 180° – 106° = 74°
From interior angles theorem, ∠ABC + ∠BCA + ∠CAB = 180°
∠BCA = 180° – 50° – 74°
= 56°
(iii)
Solution:
∠EAF = ∠BAC = 65° [vertically opposite angles]
∠ACD + ∠ACB = 180° [linear angle]
∠ACB = 180° – ∠ACD = 180° – 100° = 80°
From interior angle theorem, ∠BAC + ∠ACB + ∠ABC = 180°
∠ABC = 180° – ∠BAC – ∠ACB = 180° – 80° – 65° = 35°
(iv)
Solution:
∠ACD + ∠ACB = 180° [linear angle]
∠ACB = 180° – ∠ACD = 180° – 112° = 68°
∠EAB + ∠BAC = 180° [linear angle]
∠BAC = 180° – ∠EAB = 180° – 120° = 60°
From interior angle theorem, ∠BAC + ∠ACB + ∠ABC = 180°
∠ABC = 180° – ∠BAC – ∠ACB
x = 180° – 60° – 68°
x = 52°
(v)
Solution:
Triangle ABC is an isosceles triangle, AB = BC, then, ∠ACB = ∠BAC = 20°
From interior angle theorem, ∠BAC + ∠ACB + ∠ABC = 180°
∠ABC = 180° – ∠BAC – ∠ACB
∠ABC = 180° – 20° – 20°
∠ABC = 140°
∠ABD = x
∠ABD + ∠ABC = 180°
x = 180° – 140° = 40°
4. In the figure , QT ⏊ PR , TQR = 40˚ and SPR = 30˚ , Find TRS and PSQ.
Solution:
In Δ TRQ,
∠QTR = 90˚ ( data)
∠TQR = 40˚ (data)
∠TQR = 180˚ – ( 90˚ + 40˚ ) (Remaining angle)
= 180˚ – 130˚
∠TRQ = 50˚
In Δ PSR,
Ext ∠PSQ = ∠SPR + ∠PRS ( Exterior angle theorem)
= 30˚ + 50˚
∴Ext∠PSQ = 80˚
5. An exterior angle of triangle is 120° and one of the interior opposite angles is 30°. Find the other angles of the triangles.
Solution:
∠ACB + ∠ACD = 180˚ ( Proposition 1: Let AB be a straight line and OC be a ray standing on the line AB. Then ∠BOC + ∠COA = 180°)
∠ACB = 180˚ – 120˚
∠ACB = 60˚ ( Third angle)
In ABC,
∠A + ∠B + ∠C = 180˚
( Interior angle theorem)
∠A = 180˚- ( 30˚ + 60˚ )
∠A = 180˚ – 90˚
∠A = 90˚ ( Other interior opposite angle)
5. Find the sum of all the angles at the five vertices of the adjoining star.
Solution:
∠2 + ∠4 + ∠6 = 180˚
(Interior angle theorem)
∠1 + ∠8 + ∠10 = 180˚ ( Th .3)
∠2 + ∠9 + ∠5 = 180˚ ( Th.3)
∠1 + ∠8 + ∠4 = 180˚ ( Th.3)
∠3 + ∠7 + ∠5 = 180˚ ( Th.3)
Adding above equations,
2 ∠1 + 2 ∠2 + 2 ∠3 + 2 ∠4 + 2 ∠5 + ∠6 + ∠7 + ∠8 + ∠9 + ∠10
= 5 x 180°
2 ( ∠1 + ∠2 + ∠3 + ∠4 + ∠5 ) + ∠6 + ∠7 + ∠8 + ∠9 + ∠10 = 900°
2 ( ∠1 + ∠2 + ∠3 + ∠4 + ∠5 ) + 540˚ ( Sum of angles of a pertragon is 540˚) = 900°
2 ( ∠1 + ∠2 + ∠3 + ∠4 + ∠5 ) = 900˚ – 540˚
∠1 + ∠2 + ∠3 + ∠4 + ∠5 = 360˚/2 = 180˚
∠1 + ∠2 + ∠3 + ∠4 + ∠5 = 180˚
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