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Answers
Option (b)
Explanation:
Given :-
In ∆ PQR , < PQR = x° , < RQP = x° and
< PRQ = y°
To find :-
Find the value of x ?
Solution :-
Method -1:-
In ∆ PQR , < PQR = x° , < RQP = x° and
< PRQ = y°
The exterior angle = 140°
We know that
The sum of all the three angles in a triangle is 180°
=> < PQR + < RQP +< PRQ = 180°
=> x° + x° + y° = 180°
=> 2x° + y° = 180° ----------(1)
We know that
The exterior angle formed by extending one side of the triangle is equal to the sum of two opposite interior angles
=> 140° = x°+x°
=> 140° = 2x°
=> 2x° = 140°
=> x° = 140°/2
=> x° = 70°
Now
On Substituting the value of x in (1) then
=> 2(70°)+y° = 180°
=> 140°+y° = 180°
=> y° = 180°-140°
=> y° = 40°
Therefore, x = 70° and y = 40°
Method-2:-
In ∆ PQR , < PQR = x° , < RQP = x° and
< PRQ = y°
The exterior angle = 140°
140°+y = 180°
(linear pair)
=> y = 180°-140°
=> y = 40°
We know that
The sum of all the three angles in a triangle is 180°
=> < PQR + < RQP +< PRQ = 180°
=> x° + x° + y° = 180°
=> 2x°+40° = 180°
=> 2x° = 180°-40°
=> 2x° = 140°
=> x° = 140°/2
=> x° = 70°
Answer:-
The value of x = 70° for the given problem.
Used formulae:-
→ The sum of all the three angles in a triangle is 180° .
→ The exterior angle formed by extending one side of the triangle is equal to the sum of two opposite interior angles.
→ The sum of two adjacent angles is 180° then they are called a linear pair.
Answer:
I HAVE TO GO OUT OF STATION