Selina Solutions Concise Maths Class 7 Chapter 15 Triangles Exercise 15B are designed with the aim of improving problem solving abilities among students. The classification of triangles based on the angles and length of sides are the major concepts discussed under this exercise. Students can cross check their answers and the method of solving using the solutions created by the faculty at BYJUâ€™S. The solved examples create better conceptual knowledge, which are important from the exam perspective. Selina Solutions Concise Maths Class 7 Chapter 15 Triangles Exercise 15B, PDF links are given below for free download.

## Selina Solutions Concise Maths Class 7 Chapter 15: Triangles Exercise 15B Download PDF

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#### Exercise 15B page: 180

**1. Find the unknown angles in the given figures:**

**Solution:**

(i) From the figure (i)

x = y as the angles opposite to equal sides

In a triangle

x + y + 80^{0} = 180^{0}

Substituting the values

x + x + 80^{0} = 180^{0}

By further calculation

2x = 180^{0} â€“ 80^{0} = 100^{0}

x = 100^{0}/2 = 50^{0}

Therefore, x = y = 50^{0}.

(ii) From the figure (ii)

b = 40^{0} as the angles opposite to equal sides

In a triangle

a + b + 40^{0} = 180^{0}

Substituting the values

a + 40^{0} + 40^{0} = 180^{0}

By further calculation

a = 180 â€“ 80 = 100^{0}

Therefore, a = 100^{0} and b = 40^{0}.

(iii) From the figure (iii)

x = y as the angles opposite to equal sides

In a triangle

x + y + 90^{0} = 180^{0}

Substituting the values

x + x + 90^{0} = 180^{0}

By further calculation

2x = 180 â€“ 90 = 90^{0}

x = 90/2 = 45^{0}

Therefore, x = y = 45^{0}.

(iv) From the figure (iv)

a = b as the angles opposite to equal sides are equal

In a triangle

a + b + 80^{0 }= 180^{0}

Substituting the values

a + a + 80^{0 }= 180^{0}

By further calculation

2a = 180 â€“ 80 = 100^{0}

a = 100^{0}/2 = 50^{0}

Here a = b = 50^{0}

We know that in a triangle the exterior angle is equal to sum of its opposite interior angles

x = a + 80^{0}

So we get

x = 50 + 80 = 130^{0}

Therefore, a = 50^{0}, b = 50^{0} and x = 130^{0}.

(v) From the figure (v)

In an isosceles triangle consider each equal angle = x

x + x = 86^{0}

2x = 86^{0}

So we get

x = 86^{0}/2 = 43^{0}

For a linear pair

p + x = 180^{0}

Substituting the values

p + 43^{0} = 180^{0}

By further calculation

p = 180 â€“ 43 = 137^{0}

Therefore, p = 137^{0}.

**2. Apply the properties of isosceles and equilateral triangles to find the unknown angles in the given figures:**

**Solution:**

(i) a = 70^{0} as the angles opposite to equal sides are equal

In a triangle

a + 70^{0} + x = 180^{0}

Substituting the values

70^{0} + 70^{0} + x = 180^{0}

By further calculation

x = 180 â€“ 140 = 40^{0}

y = b as the angles opposite to equal sides are equal

Here a = y + b as the exterior angle is equal to sum of interior opposite angles

70^{0} = y + y

So we get

2y = 70^{0}

y = 70^{0}/2 = 35^{0}

Therefore, x = 40^{0} and y = 35^{0}.

(ii) From the figure (ii)

Each angle is 60^{0} in an equilateral triangle

In an isosceles triangle

Consider each base angle = a

a + a + 100^{0} = 180^{0}

By further calculation

2a = 180 â€“ 100 = 80^{0}

So we get

a = 80^{0}/2 = 40^{0}

x = 60^{0} + 40^{0} = 100^{0}

y = 60^{0} + 40^{0} = 100^{0}

(iii) From the figure (iii)

130^{0} = x + p as the exterior angle is equal to the sum of interior opposite angles

It is given that the lines are parallel

Here p = 60^{0} is the alternate angles and y = a

In a linear pair

a + 130^{0} = 180^{0}

By further calculation

a = 180 â€“ 130 = 50^{0}

Here x + p = 130^{0}

Substituting the values

x + 60^{0} = 130^{0}

By further calculation

x = 130 â€“ 60 = 70^{0}

Therefore, x = 70^{0}, y = 50^{0} and p = 60^{0}.

(iv) From the figure (iv)

x = a + b

Here b = y and a = c as the angles opposite to equal sides are equal

a + c + 30^{0} = 180^{0}

Substituting the values

a + a + 30^{0} = 180^{0}

By further calculation

2a = 180 â€“ 30 = 150^{0}

a = 150/2 = 75^{0}

We know that

b + y = 90^{0}

Substituting the values

y + y = 90^{0}

2y = 90^{0}

y = 90/2 = 45^{0}

where b = 45^{0}

Therefore, x = a + b = 75 + 45 = 120^{0} and y = 45^{0}.

(v) From the figure (v)

a + b + 40^{0} = 180^{0}

So we get

a + b = 180 â€“ 40 = 140^{0}

The angles opposite to equal sides are equal

a = b = 140/2 = 70^{0}

x = b + 40^{0} = 70^{0}Â + 40^{0}= 110^{0}

Here the exterior angle of a triangle is equal to the sum of its interior opposite angles

In the same way

y = a + 40^{0}

Substituting the values

y = 70^{0}Â + 40^{0} = 110^{0}

Therefore, x = y = 110^{0}.

**3. The angle of vertex of an isosceles triangle is 100Â°. Find its base angles.**

**Solution:**

Consider âˆ† ABC

Here AB = AC and âˆ B = âˆ C

We know that

âˆ A = 100^{0}

In a triangle

âˆ A + âˆ B + âˆ C = 180^{0}

Substituting the values

100^{0} + âˆ B + âˆ B = 180^{0}

By further calculation

2âˆ B = 180^{0} â€“ 100^{0} = 80^{0}

âˆ B = 80/2 = 40^{0}

Therefore, âˆ B = âˆ C = 40^{0}.

**4. One of the base angles of an isosceles triangle is 52Â°. Find its angle of vertex.**

**Solution:**

It is given that the base angles of isosceles triangle ABC = 52^{0}

Here âˆ B = âˆ C = 52^{0}

In a triangle

âˆ A + âˆ B + âˆ C = 180^{0}

Substituting the values

âˆ A + 52^{0} + 52^{0} = 180^{0}

By further calculation

âˆ A = 180 â€“ 104 = 76^{0}

Therefore, âˆ A = 76^{0}.

**5. In an isosceles triangle, each base angle is four times of its vertical angle. Find all the angles of the triangle.**

**Solution:**

Consider the vertical angle of an isosceles triangle = x

So the base angle = 4x

In a triangle

x + 4x + 4x = 180^{0}

By further calculation

9x = 180^{0}

x = 180/9 = 20^{0}

So the vertical angle = 20^{0}

Each base angle = 4x = 4 Ã— 20^{0} = 80^{0}

**6. The vertical angle of an isosceles triangle is 15Â° more than each of its base angles. Find each angle of the triangle.**

**Solution:**

Consider the angle of the base of isosceles triangle = x^{0}

So the vertical angle = x + 15^{0}

In a triangle

x + x + x + 15^{0} = 180^{0}

By further calculation

3x = 180 â€“ 15 = 165^{0}

x = 165/3 = 55^{0}

Therefore, the base angle = 55^{0}

Vertical angle = 55 + 15 = 70^{0}.

**7. The base angle of an isosceles triangle is 15Â° more than its vertical angle. Find its each angle.**

**Solution:**

Consider the vertical angle of the isosceles triangle = x^{0}

Here each base angle = x + 15^{0}

In a triangle

x + 15^{0} + x + 15^{0}Â + x = 180^{0}

By further calculation

3x + 30^{0} = 180^{0}

3x = 180 â€“ 30 = 150^{0}

x = 150/3 = 50^{0}

Therefore, vertical angle = 50^{0} and each base angle = 50 + 15 = 65^{0}.

**8. The vertical angle of an isosceles triangle is three times the sum of its base angles. Find each angle.**

**Solution:**

Consider each base angle of an isosceles triangle = x

Vertical angle = 3 (x + x) = 3 (2x) = 6x

In a triangle

6x + x + x = 180^{0}

By further calculation

8x = 180^{0}

x = 180/8 = 22.5^{0}

Therefore, each base angle = 22.5^{0} and vertical angle = 3 (22.5 + 22.5) = 3 Ã— 45 = 135^{0}.

**9. The ratio between a base angle and the vertical angle of an isosceles triangle is 1 : 4. Find each angle of the triangle.**

**Solution:**

It is given that the ratio between a base angle and the vertical angle of an isosceles triangle = 1: 4

Consider base angle = x

Vertical angle = 4x

In a triangle

x + x + 4x = 180^{0}

By further calculation

6x = 180^{0}

x = 180/6 = 30^{0}

Therefore, each base angle = x = 30^{0} and vertical angle = 4x = 4 Ã— 30^{0} = 120^{0}.

**10. In the given figure, BI is the bisector of âˆ ABC and CI is the bisector of âˆ ACB. Find âˆ BIC.**

**Solution:**

In âˆ† ABC

BI is the bisector of âˆ ABC and CI is the bisector of âˆ ACB

Here AB = AC

âˆ B = âˆ C as the angles opposite to equal sides are equal

We know that âˆ A = 40^{0}

In a triangle

âˆ A + âˆ B + âˆ C = 180^{0}

Substituting the values

40^{0} + âˆ B + âˆ B = 180^{0}

By further calculation

40^{0}Â + 2âˆ B = 180^{0}

2âˆ B = 180 â€“ 40 = 140^{0}

âˆ B = 140/2 = 70^{0}

Here BI and CI are the bisectors of âˆ ABC and âˆ ACB

âˆ IBC = Â½ âˆ ABC = Â½ Ã— 70^{0} = 35^{0}

âˆ ICB = Â½ âˆ ACB = Â½ Ã— 70^{0} = 35^{0}

In âˆ† IBC

âˆ BIC + âˆ IBC + âˆ ICB = 180^{0}

Substituting the values

âˆ BIC + 35^{0} + 35^{0}Â = 180^{0}

By further calculation

âˆ BIC = 180 â€“ 70 = 110^{0}

Therefore, âˆ BIC = 110^{0}.

**11. In the given figure, express a in terms of b.**

**Solution:**

From the âˆ† ABC

BC = BA

âˆ BCA = âˆ BAC

Here the exterior âˆ CBE = âˆ BCA + âˆ BAC

a = âˆ BCA + âˆ BCA

a = 2âˆ BCA â€¦â€¦ (1)

Here âˆ ACB = 180^{0} â€“ b

Where âˆ ACD and âˆ ACB are linear pair

âˆ BCA = 180^{0} â€“ b â€¦â€¦. (2)

We get

a = 2 âˆ BCA

Substituting the values

a = 2 (180^{0} â€“ b)

a = 360^{0} â€“ 2b

**12. (a) In Figure (i) BP bisects âˆ ABC and AB = AC. Find x.
(b) Find x in Figure (ii) Given: DA = DB = DC, BD bisects âˆ ABC and âˆ ADB = 70Â°.**

**Solution:**

(a) From the figure (i)

AB = AC and BP bisects âˆ ABC

AP is drawn parallel to BC

Here PB is the bisector of âˆ ABC

âˆ PBC = âˆ PBA

âˆ APB = âˆ PBC are alternate angles

x = âˆ PBC â€¦.. (1)

In âˆ† ABC

âˆ A = 60^{0}

Since AB = AC we get âˆ B = âˆ C

In a triangle

âˆ A + âˆ B + âˆ C = 180^{0}

Substituting the values

60^{0} + âˆ B + âˆ C = 180^{0}

We get

60^{0} + âˆ B + âˆ B = 180^{0}

By further calculation

2âˆ B = 180 â€“ 60 = 120^{0}

âˆ B = 120/2 = 60^{0}

Â½ âˆ B = 60/2 = 30^{0}

âˆ PBC = 30^{0}

So from figure (i) x = 30^{0}

(b) From the figure (ii)

DA = DB = DC

Here BD bisects âˆ ABC and âˆ ADB = 70^{0}

In a triangle

âˆ ADB + âˆ DAB + âˆ DBA = 180^{0}

Substituting the values

70^{0} + âˆ DBA + âˆ DBA = 180^{0}

By further calculation

70^{0} + 2âˆ DBA = 180^{0}

2âˆ DBA = 180 â€“ 70 = 110^{0}

âˆ DBA = 110/2 = 55^{0}

Here BD is the bisector of âˆ ABC

So âˆ DBA = âˆ DBC = 55^{0}

In âˆ† DBC

DB = DC

âˆ DCB = âˆ DBC

Hence, x = 55^{0}.

**13. In each figure, given below, ABCD is a square and âˆ† BEC is an equilateral triangle.
**

**Find, in each case: (i) âˆ ABE (ii) âˆ BAE**

**Solution:**

The sides of a square are equal and each angle is 90^{0}

In an equilateral triangle all three sides are equal and all angles are 60^{0}

In figure (i) ABCD is a square and âˆ† BEC is an equilateral triangle

(i) âˆ ABE = âˆ ABC + âˆ CBE

Substituting the values

âˆ ABE = 90^{0} + 60^{0}= 150^{0}

(ii) In âˆ† ABE

âˆ ABE + âˆ BEA + âˆ BAE = 180^{0}

Substituting the values

150^{0} + âˆ BAE + âˆ BAE = 180^{0}

By further calculation

2âˆ BAE = 180 â€“ 150 = 30^{0}

âˆ BAE = 30/2 = 15^{0}

In figure (ii) ABCD is a square and âˆ† BEC is an equilateral triangle

(i) âˆ ABE = âˆ ABC – âˆ CBE

Substituting the values

âˆ ABE = 90^{0} â€“ 60^{0} = 30^{0}

(ii) In âˆ† ABE

âˆ ABE + âˆ BEA + âˆ BAE = 180^{0}

Substituting the values

30^{0} + âˆ BAE + âˆ BAE = 180^{0}

By further calculation

2âˆ BAE = 180 â€“ 30 = 150^{0}

âˆ BAE = 150/2 = 75^{0}

**14. In âˆ† ABC, BA and BC are produced. Find the angles a and h. if AB = BC.**

**Solution:**

In âˆ† ABC, BA and BC are produced

âˆ ABC = 54^{0} and AB = BC

In âˆ† ABC

âˆ BAC + âˆ BCA + âˆ ABC = 180^{0}

Substituting the values

âˆ BAC + âˆ BAC + 54^{0 }= 180^{0}

2âˆ BAC = 180 â€“ 54 = 126^{0}

âˆ BAC = 126/2 = 63^{0}

âˆ BCA = 63^{0}

In a linear pair

âˆ BAC + b = 180^{0}

Substituting the value

63^{0} + b = 180^{0}

So we get

b = 180 â€“ 63 = 117^{0}

In a linear pair

âˆ BCA + a = 180^{0}

Substituting the value

63^{0}Â + a = 180^{0}

So we get

a = 180 â€“ 63 = 117^{0}

Therefore, a = b = 117^{0}.