Test: Angles Sum Property A Triangle - Class 9 MCQ

so interior angle corresponding to 80° will be 100°
let there ratio be x so angles would be x and 3x
x + 3x + 100° = 180°
4x = 80°
x = 20°
so x and 3x  will be 20° and 60° respectively  

interior ∠A = 180 - 110 = 70
interior ∠C = 180 - 120 = 60
interior ∠A + interior ∠C + x =180
70 + 60 + x = 180
130 + x =180
x = 50
 

To find angle c:

c = 180° - (20° + 62°) = 98°
(Angle sum property of the triangle)

Therefore, angle a = 98°. When lines are parallel, alternate interior angles are equal.

Solution:

- Given that in triangle ABC, AB = AC
- We know that in an isosceles triangle, the base angles are equal
- So, angle B = angle C
- Let the value of angle B be r
- Therefore, angle C = r
- From angle sum property of a triangle, we have: A + B + C = 180°
- Substituting the values, we get: A + r + r = 180°
- Simplifying, we get: A + 2r = 180°
- 80° +2r =180°
- 2r = 100°
- r = 50°
so x + r = 180°
    x + 50° = 180°
    x = 130°

Let us consider the sum of two angles as ∠A+∠B=116∘ and the difference can be written as ∠A−∠B=24^o

We know that the sum of all the angles in a triangle is 180^o.

So we can write it as

∠A+∠B+∠C=180^o

By substituting ∠A+∠B=116^o in the above equation
116^o + ∠C = 180^o

On further calculation

∠C=180^o − 116^o

By subtraction

∠C=64^o

It is given that ∠A−∠B=24^o

It can be written as ∠A=24^o+∠B

Now by substituting ∠A=24^o+∠Bin∠A+∠B=116^o

∠A+∠B=116^o

24^o+∠B+∠B=116^o

On further calculation

24^o+2∠B=116^o

By subtraction

2∠B=116^o−24^o

2∠B=92^o

By division

∠B=292​
∠B=46∘
By substituting ∠B=46∘ in ∠A=24∘+∠B

We get

∠A=24^o+46^o
By addition
∠A=70^o
Therefore, ∠A=70^o,∠B=46^o and ∠C=64^o