**GSEB Solutions for Class 9 Mathematics – Triangle (English Medium)**

**Exercise 9:**

**Solution 1:**

In ∆ABC,

m∠A + m∠B + m∠C = 180°

∴100° + m∠C = 180° … (∵ m∠A + m∠B = 100°)

∴ m∠C = 180° – 100°

∴ m∠C = 80°

Given, m∠B + m∠C = 130°

∴ m∠B + 80° = 130°

∴ m∠B = 130° – 80°

∴ m∠B = 50°

Now, m∠A + m∠B = 100°

∴ m∠A + 50° = 100°∴ m∠A = 100° – 50°

∴ m∠A = 50°

In ∆ABC, m∠A = 50°, m∠B = 50° and m∠C = 80°.

**Solution 2:**

**Solution 3:**

The length of each side of a triangle is greater than the positive difference of the lengths of the other two sides.

In ∆ABC,

AC < AB + BC and AC > |AB – BC|

∴|AB – BC| < AC < AB + BC

∴|10 – 18| < AC < 10 + 18

∴|-8| < AC < 28

∴8 < AC < 28

**Solution 4:**

**Solution 5:**

**Solution 6:**

**Solution 7:**

The length of each side of a triangle is greater than the positive difference of the lengths of the other two sides.

In ΔABC, AB < BC + AC and AB > |BC – AC|

∴|BC – AC| < AB < BC + AC

∴|5 – 12| < AB < 5 + 12

∴|-7| < AB < 17

∴7 < AB < 17

**Solution 8:**

**Solution 9:**

**Solution 10:**

The length of each side of a triangle is greater than the positive difference of the lengths of the other two sides.

In DABC, AC < AB + BC and AC > |AB – BC|

∴|AB – BC| < AC < AB + BC

∴|8 – 5| < AC < 8 + 5

∴|3| < AB < 13

∴3 < AB < 13

**Solution 11(1):**

**Solution 11(2):**

**Solution 11(3):**

b. 75°

∠ACD is an exterior angle of ∆ ABC.

∴∠ ACD and ∠ ACB is linear pair of angle.

∴m∠ ACD + m∠ ACB = 180°

∴ 105° + m∠ ACB = 180°

∴ m∠ ACB = 180° – 105°

∴ m∠ ACB = 75°

**Solution 11(4):**

c. C

For the corresponding BAC ↔ YXZ between ∆ ABC and ∆ XYZ, the angle ∠C corresponds to ∠Z.

**Solution 11(5):**

**Solution 11(6):**

**Solution 11(7):**

a. 10

In ∆ ABC,

Given,

BC = 3, AC = 4

∠A ≅ ∠C

∴BC = AB

The perimeter of ∆ABC = AB + BC + AC = 3 + 3 + 4 = 10

**Solution 11(8):**

**Solution 11(9):**

c. AAA

AAA – condition is not possible for the congruence of two triangles.

**Solution 11(10):**

d. AC < AB + BC

For ∆ABC, AC < AB + BC is true (if it is not a right triangle)

**Solution 11(11):**

**Solution 11(12):**

**Solution 11(13):**

**Solution 11(14):**

a. scalene

In ∆ ABC,

m∠A + m∠B + m∠C = 180°

x + 2x + y = 180°

∴3x + y = 180° …(1)

∴2x – y = 40° … (2)

Add (1) and (2).

∴5x = 220°

∴x = 44°

Substitute in(1) we get,

3(44) + y = 180°

∴y = 48°

**Solution 11(15):**

a. 30°

Let m∠ A = k, m∠ B = 2k, m∠ C = 3k

In ∆ ABC

m∠ A : m∠ B : m∠ C = 1 : 2 : 3

∴m∠ A + m∠ B + m∠ C = 180°

∴k + 2k + 3k = 180°

∴6k = 180°

∴k = 30°

m∠A = 30°, m∠B = 60°, m∠C = 90°

The measure of the smallest angle is 30°.

**Solution 11(16):**

**Solution 11(17):**

c. 60°

In ∆ ABC,

m∠A + m∠B + m∠C = 180°

∴ 120° + m∠C = 180°

∴ m∠C = 180° – 120° = 60°

**Exercise 9.1:**

**Solution 1:**

In ∆ABC,

∠ACD is an exterior angle and m∠ACD = 120.

The measure of one of the interior opposite angle is 40.

Let m∠A = 40.

We know that, measure of any exterior angle of a triangle is equal to the sum of the measures of its two interior opposite angles.

m∠ACD = m∠A + m∠B

∴ 120 = 40 + m∠B

∴ m∠B = 120 – 40= 80

In DABC,

The sum of the measures of all three angles of a triangle is 180.

m∠A + m∠B + m∠C = 180

∴ 40 + 80 + m∠C = 180

∴ m∠C = 180 – 40 – 80 = 60

Thus, the measures of the remaining angles of ∆ABC are 80 and 60.

**Solution 2:**

**Solution 3:**

Let in ∆PQR,

m∠P : m∠Q : m∠R = 2 : 3 : 5

Then,

m∠P = 2k, then m∠Q = 3k and m∠R= 5k.

In ∆PQR,

The sum of the measures of all three angles of triangle is 180.

m∠P : m∠Q : m∠R = 180

∴ 2k + 3k + 5k = 180

∴ 10k = 180

∴ k = 18

Substitute k = 18,we get

m∠P = 2k = 2 × 18 = 36

m∠Q = 3k = 3 × 18 = 54

m∠R = 5k = 5 × 18 = 90

Thus, the measures of the angles of ∆ABC are 36, 54 and 90.

**Solution 4:**

Figure 1:

M ∠BAC + m ∠BAE = 180 (linear pair)

∴ m ∠BAC + 110 = 180

∴ m ∠BAC = 180 – 110 = 70

M ∠ACB + m ∠ACD = 180 (linear pair)

∴ m ∠ACB + 105 = 180

∴ m ∠ACB = 180 – 105

∴ m ∠ACB = 75

In DABC,

The sum of the measures of all three angles of triangle is 180.

∴ m ∠BAC + m ∠ABC + m ∠ACB = 180

∴ 70 + x + 75 = 180

∴ x = 180 – 70 – 75 = 35

Figure 2 :

We know that, measure of any exterior angle of a triangle is equal to the sum of the measure of its two interior opposite angles.

m∠ABE = m∠A + m∠ACB

∴ 105 = m∠A + m∠ACB ………….. (1)

m∠ACD = m∠A + m∠ABC

∴ 100 = m∠A + m∠ABC ………….. (2)

Add (1) and (2), we get

105 + 100 = m∠A + m∠ACB + m∠A + m∠ABC

∴ 205 = m∠A + (m∠ACB + m∠A + m∠ABC)

[The sum of the measures of all three angles of triangle is 180. ∴ m∠ACB + m∠A + m∠ABC = 180]

∴ 205 = m∠A + 180

∴ m∠A = 205 – 180

∴ m∠A = 25|

∴ x = 25

**Solution 5:**

Let, m∠B = x

m∠A – m∠B = 70

∴ m∠A – x = 70

∴ m∠A = x + 70

m∠B – m∠C = 40

∴ x – m∠C = 40

∴ x – 40 = m∠C

∴ m∠C = x – 40

In ∆ABC,

m∠A + m∠B + m∠C = 180 (The sum of the measures of all three angles of triangle is 180)

∴x + 70 + x + x – 40 = 180

∴ 3x + 30 = 180

∴ 3x = 180 – 30

∴ 3x = 150

∴ x = 50

m∠A = x + 70 = 50 + 70 = 120

m∠B = x = 50

m∠C = x – 4 = 50 – 40 = 10

Thus, m∠A = 120 and m∠C = 10.

**Solution 6:**

**Solution 7:**

**Solution 8:**

We know that, measure of any exterior angle of a triangle is equal to the sum of the measure of its two interior opposite angles.

Given, m∠ABE is an exterior angle of DABC.

∴m∠ABE = m∠BAC + m∠ACB …[1]

∠CAD is an exterior angle of ∆ABC.

∴m∠CAD = m∠ABC + m∠ACB …[2]

Add equation [1] and [2], we get

m∠ABE + m∠CAD = m∠BAC + m∠ACB + m∠ABC + m∠ACB

∴ 100 + 125 = (m∠BAC + m∠ACB + m∠ABC) + m∠ACB

∴ 225 = 180 + m∠ACB (The sum of the measures of all three angles of triangle is 180.)

∴ m∠ACB = 225 – 180 = 45

**Exercise 9.2:**

**Solution 1:**

**Solution 2:**

**Solution 3:**

**Solution 4:**

**Exercise 9.3:**

**Solution 1:**

**Solution 2:**

**Solution 3:**

**Solution 4:**

**Solution 5:**

**Solution 6:**

**Solution 7:**

**Solution 8:**

**Solution 9:**

**Solution 10:**

Let the measure of two congruent angles of the triangles are x and x.

The measure of the third angle is x + 60.

Sum of measure of all three angles of triangle is 180°.

∴x + x + (x + 60) = 180°

∴3x + 60 = 180

∴3x = 180 – 60

∴3x = 120

∴x = 40

Then, x + 60 = 40 + 60 = 100

The measures of the two congruent angles of the triangle are 40° each and the measure of the third angle is 100°.

**Exercise 9.4:**

**Solution 1:**

**Solution 2:**

**Solution 3:**

**Solution 4:**

**Solution 5:**

**Solution 6:**

Given: P is in the interior of DABC such that PA = PB = PC.

To prove: m∠A = m∠ABP + m∠ACP.

Proof : In ∆ABP,

PA = PB (Given)

∴ m∠ABP = m∠BAP …….. (1)

In ∆APC, PA = PC (data)

∴ m∠ACP = m∠PAC …….. (2)

Adding (1) and (2), we get

m∠ABP + m∠ACP = m∠BAP + m∠PAC

Now, P is in the interior of ∆ABC.

m∠BAP + m∠PAC = m∠BAC

m∠ABP + m∠ACP = m∠BAC

⇒ m∠ABP + m∠ACP = m∠A

m∠A = m∠ABP + m∠ACP.

**Solution 7:**

**Exercise 9.5:**

**Solution 1:**

**Solution 2:**

**Solution 3:**

**Solution 4:**

**Solution 5:**

**Solution 6:**