Chapter 6 Exercise 6.2 Page: 118 - Wisdom TechSavvy Academy

Online Class For 7th Standard Students (CBSE)

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Solution:-

We Know That,

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

= x = 65^o+ 45^o

= x = 110^o

We Know That,

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

= x = 60^o+ 60^o

= x = 120^o

Solution:-

We Know That,

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

= x = 30^o+ 60^o

= x = 90^o

2. Find the value of the unknown interior angle x in the following figures:

(i)

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

= 70^o + x = 100^o

By transposing 70^o from LHS to RHS it becomes – 70^o

= x = 100^o – 70^o

= x = 30^o

(iii)

Solution:-

We Know That,

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

= x + 60^o = 120^o

By transposing 60^o from LHS to RHS it becomes – 60^o

= x = 120^o – 60^o

= x = 60^o

(v)

Solution:-

We Know That,

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

The given triangle is a right angled triangle. So the angle opposite to the x is 90^o.

= x + 35^o = 75^o

By transposing 35^o from LHS to RHS it becomes – 35^o

= x = 75^o – 35^o

= x = 40^o