Selina Solutions Concise Maths Class 7 Chapter 15 Triangles Exercise 15A consists of problems on determining the angle of the given set of triangles. A triangle is a closed figure which has three sides and vertices. The main aim of creating the exercise wise solutions is to provide a strong foundation of fundamental concepts covered in this exercise. These concepts are important as it will be continued in further classes also. For a clear conceptual knowledge, students can access Selina Solutions Concise Maths Class 7 Chapter 15 Triangles Exercise 15A PDF, from the links which are available here.
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Exercise 15A page: 176
1. State, if the triangles are possible with the following angles :
(i) 20°, 70° and 90°
(ii) 40°, 130° and 20°
(iii) 60°, 60° and 50°
(iv) 125°, 40° and 15°
Solution:
In a triangle, the sum of three angles is 1800
(i) 20°, 70° and 90°
Sum = 200 + 700 + 900 = 1800
Here the sum is 1800 and therefore it is possible.
(ii) 40°, 130° and 20°
Sum = 40° + 130° + 20° = 1900
Here the sum is not 1800 and therefore it is not possible.
(iii) 60°, 60° and 50°
Sum = 60° + 60° + 50° = 1700
Here the sum is not 1800 and therefore it is not possible.
(iv) 125°, 40° and 15°
Sum = 125° + 40° + 15° = 1800
Here the sum is 1800 and therefore it is possible.
2. If the angles of a triangle are equal, find its angles.
Solution:
In a triangle, the sum of three angles is 1800
So each angle = 1800/3 = 600
3. In a triangle ABC, ∠A = 45° and ∠B = 75°, find ∠C.
Solution:
In a triangle, the sum of three angles is 1800
∠A + ∠B + ∠C = 1800
Substituting the values
450 + 750 + ∠C = 1800
By further calculation
1200 + ∠C = 1800
So we get
∠C = 1800 – 1200 = 600
4. In a triangle PQR, ∠P = 60° and ∠Q = ∠R, find ∠R.
Solution:
Consider ∠Q = ∠R = x
∠P = 60°
We can write it as
∠P + ∠Q + ∠R = 1800
Substituting the values
600 + x + x = 1800
By further calculation
600 + 2x = 1800
2x = 1800 – 600 = 1200
So we get
x = 1200/2 = 600
∠Q = ∠R = 600
Therefore, ∠R = 600.
5. Calculate the unknown marked angles in each figure:
Solution:
In a triangle, the sum of three angles is 1800
(i) From figure (i)
900 + 300 + x = 1800
By further calculation
1200 + x = 1800
So we get
x = 1800 – 1200 = 600
Therefore, x = 600.
(ii) From figure (ii)
y + 800 + 200 = 1800
By further calculation
y + 1000 = 1800
So we get
y = 1800 – 1000 = 800
Therefore, y = 800.
(iii) From figure (iii)
a + 900 + 400 = 1800
By further calculation
a + 1300 = 1800
So we get
a = 1800 – 1300 = 500
Therefore, a = 500.
6. Find the value of each angle in the given figures:
Solution:
(i) From the figure (i)
∠A + ∠B + ∠C = 1800
Substituting the values
5x0 + 4x0 + x0 = 1800
By further calculation
10x0 = 1800
x = 180/10 = 180
So we get
∠A = 5x0 = 5 × 180 = 900
∠B = 4x0 = 4 × 180 = 720
∠C = x = 180
(ii) From the figure (ii)
∠A + ∠B + ∠C = 1800
Substituting the values
x0 + 2x0 + 2x0 = 1800
By further calculation
5x0 = 1800
x0 = 1800/5 = 360
So we get
∠A = x0 = 360
∠B = 2x0 = 2 × 360 = 720
∠C = 2x0 = 2 × 360 = 720
7. Find the unknown marked angles in the given figure:
Solution:
(i) From the figure (i)
∠A + ∠B + ∠C = 1800
Substituting the values
b0 + 500 + b0 = 1800
By further calculation
2b0 = 1800 – 500 = 1300
b0 = 1300/2 = 650
Therefore, ∠A = ∠C = b0 = 650.
(ii) From the figure (ii)
∠A + ∠B + ∠C = 1800
Substituting the values
x0 + 900 + x0 = 1800
By further calculation
2x0 = 1800 – 900 = 900
x0 = 900/2 = 450
Therefore, ∠A = ∠C = x0 = 450.
(iii) From the figure (iii)
∠A + ∠B + ∠C = 1800
Substituting the values
k0 + k0 + k0 = 1800
By further calculation
3k0 = 1800
k0 = 1800/3 = 600
Therefore, ∠A = ∠B = ∠C = 600.
(iv) From the figure (iv)
∠A + ∠B + ∠C = 1800
Substituting the values
(m0 – 50) + 600 + (m0 + 50) = 1800
By further calculation
m0 – 50 + 600 + m0 + 50 = 1800
2m0 = 1800 – 600 = 1200
m0 = 1200/2 = 600
Therefore, ∠A = m0 – 50 = 600 – 50 = 550
∠C = m0 + 50 = 600 + 50 = 650
8. In the given figure, show that: ∠a = ∠b + ∠c
(i) If ∠b = 60° and ∠c = 50°; find ∠a.
(ii) If ∠a = 100° and ∠b = 55°; find ∠c.
(iii) If ∠a = 108° and ∠c = 48°; find ∠b.
Solution:
From the figure
AB || CD
b = ∠C and ∠A = c are alternate angles
In triangle PCD
Exterior ∠APC = ∠C + ∠D
a = b + c
(i) If ∠b = 60° and ∠c = 50°
∠a = ∠b + ∠c
Substituting the values
∠a = 60 + 50 = 1100
(ii) If ∠a = 100° and ∠b = 55°
∠a = ∠b + ∠c
Substituting the values
∠c = 100 – 55 = 450
(iii) If ∠a = 108° and ∠c = 48°
∠a = ∠b + ∠c
Substituting the values
∠b = 108 – 48 = 600
9. Calculate the angles of a triangle if they are in the ratio 4 : 5 : 6.
Solution:
In a triangle, the sum of angles of a triangle is 1800
∠A + ∠B + ∠C = 1800
It is given that
∠A: ∠B: ∠C = 4: 5: 6
Consider ∠A = 4x, ∠B = 5x and ∠C = 6x
Substituting the values
4x + 5x + 6x = 1800
By further calculation
15x = 1800
x = 1800/15 = 120
So we get
∠A = 4x = 4 × 120 = 480
∠B = 5x = 5 × 120 = 600
∠C = 6x = 6 × 120 = 720
10. One angle of a triangle is 60°. The, other two angles are in the ratio of 5 : 7. Find the two angles.
Solution:
From the triangle ABC
Consider ∠A = 600, ∠B: ∠C = 5:7
In a triangle
∠A + ∠B + ∠C = 1800
Substituting the values
600 + ∠B + ∠C = 1800
By further calculation
∠B + ∠C = 1800 – 600 = 1200
Take ∠B = 5x and ∠C = 7x
Substituting the values
5x + 7x = 1200
12x = 1200
x = 1200/12 = 100
So we get
∠B = 5x = 5 × 100 = 500
∠C = 7x = 7 × 100 = 700
11. One angle of a triangle is 61° and the other two angles are in the ratio 1 ½ : 1 1/3. Find these angles.
Solution:
From the triangle ABC
Consider ∠A = 610
In a triangle
∠A + ∠B + ∠C = 1800
Substituting the values
610 + ∠B + ∠C = 1800
By further calculation
∠B + ∠C = 1800 – 610 = 1190
∠B: ∠C = 1 ½: 1 1/3 = 3/2: 4/3
Taking LCM
∠B: ∠C = 9/6: 8/ 6
∠B: ∠C = 9: 8
Consider ∠B = 9x and ∠C = 8x
Substituting the values
9x + 8x = 1190
17x = 1190
x = 1190/ 17 = 70
So we get
∠B = 9x = 9 × 70 = 630
∠C = 8x = 8 × 70 = 560
12. Find the unknown marked angles in the given figures:
Solution:
In a triangle, if one side is produced
Exterior angle is the sum of opposite interior angles
(i) From the figure (i)
1100 = x0 + 300
By further calculation
x0 = 1100 – 300 = 800
(ii) From the figure (ii)
1200 = y0 + 600
By further calculation
y0 = 1200 – 600 = 600
(iii) From the figure (iii)
1220 = k0 + 350
By further calculation
k0 = 1220 – 350 = 870
(iv) From the figure (iv)
1350 = a0 + 730
By further calculation
a0 = 1350 – 730 = 620
(v) From the figure (v)
1250 = a + c …… (1)
1400 = a + b …… (2)
By adding both the equations
a + c + a + b = 1250 + 1400
On further calculation
a + a + b + c = 2650
We know that a + b + c = 1800
Substituting it in the equation
a + 1800 = 2650
So we get
a = 265 – 180 = 850
If a + b = 1400
Substituting it in the equation
850 + b = 1400
So we get
b = 140 – 85 = 550
If a + c = 1250
Substituting it in the equation
850 + c = 1250
So we get
c = 125 – 85 = 400
Therefore, a = 850, b = 550 and c = 400.
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