Chapter 9 Triangles and its Angles RD Sharma Solutions Exercise 9.1 Class 9 Maths
Chapter Name  RD Sharma Chapter 9 Triangles and its Angles Exercise 9.1 
Book Name  RD Sharma Mathematics for Class 10 
Other Exercises 

Related Study  NCERT Solutions for Class 10 Maths 
Exercise 9.1 Solutions
1. In a Î”ABC, if ∠A = 55°, ∠B = 40° , find ∠C.
Solution
Given ∠A = 55° , ∠B = 40 then ∠C = ?
We know that
In a Î”ABC sum of all angles of triangle is 180°
i.e., ∠A + ∠B + ∠C = 180°
⇒ 55° + 40° + ∠C = 180°
⇒ 95° + ∠C = 180°
⇒ ∠C = 180°  95°
⇒ ∠C = 85°
2. If the angles of a triangles are in the ratio 1 : 2 : 3, determine three angles.
Solution
Given that the angles of a triangle are in the ratio 1 : 2 : 3
Let the angles be a, 2a, 3a
∴ We know that
Sum of all angles of triangles is 80°
a + 2a + 3a = 180°
⇒ 6a = 180°
⇒ a = 180°/6
⇒ a = 30°
Since a = 30°
2a = 2(30°) = 60°
3a = 3(30°) = 90°
∴ angles are a = 30°, 2a = 60°, 3a = 90°
∴ Hence angles are 30°, 60° and 90°
3. The angles of a triangle are (x  40)° , (x  20)° and (x/2  10)° . Find the value of x.
Solution
Given that
The angles of a triangle are
(x  40)° , (x  20)° and (x/2  10)°
We know that
Sum of all angles of triangle is 180°
4. The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angles is 10° , find the three angles.
Solution
Given that,
The difference between two consecutive angles is 10° .
Let x, x + 10, x + 20 be the consecutive angles differ by 10°
W.K.T sum of all angles of triangle is 180° .
x + x + 10 + x + 20 = 180°
3x + 30 = 180°
⇒ 3x = 180  30°
⇒ 3x = 150°
⇒ x = 50°
x = 50°
The required angles are
x, x + 10 and x + 20
x = 50°
⇒ x + 10 = 50 + 10 = 60
⇒ x + 20 = 50 + 10 + 10 = 70
The difference between two consecutive angles is 10° then three angles are 50°, 60° and 70°.
5. Two angles of a triangle are equal and the third angle is greater than each of those angles by 30°. Determine all the angles of the triangle.
Solution
Given that,
Two angles are equal and the third angle is greater than each of those angles by 30° .
Let x, x, x + 30 be the angles of a triangle
We know that
Sum of all angles of a triangle is 180°
x + x + x + 30 = 180°
3x + 30 = 180°
⇒ 3x = 180°  30°
⇒ 3x = 150°
⇒ x = 150°/3
⇒ x = 50°
∴ The angles are x, x, x + 30
x = 50°
⇒ x + 30 = 80°
∴ The required angles are 50°, 50°, 80°.
Now, in Î”ABC
∠A + ∠B + ∠C = 180°
⇒ ∠B + ∠B = 180° [∵ ∠B = ∠A + ∠C]
⇒ ∠B = 180°/2
i.e., ∠BOC = 135°
But, W.K.T