Question

Two angle of quadrilateral

Answer 1

given 2 angles are 75°

let angle 1 and angle 3 = 75° [given]

also, two angles are equal

so let the 2 equal angles be x

now, 75° + x + 75° + x = 360° (since it is a quadrilateral)

150° + 2x = 360°

=> 2x = 360° - 150° = 210°

=> x = 210°/ 2 = 105°

so the other 2 angles of the quadrilateral are 105°

hope it helps u mate

Answer 2

1) Given two angles are of 75° and other two are equal

Let the equal angles be x so we have

75 + 75 + x + x = 360° (Angle sum property of a quadrilateral)

=> 150 + 2x = 360°

=> 2x = 360 - 150 = 210°

=> x = 210/2

=> x = 105°

So each angles measure 105°

2) Given two angles are 100° and 50°

So by angle sum property of quadrilateral we have

A + B + 100° + 50° = 360°

=> A + B = 360° - 150° = 210°


Now given their bisector meets at P

so bisector angle sum
= A/2 + B/2
=> (A + B)/2
=> 210/2
=> 105°

This means bisected A + bisected B = 105°

So we can see a ∆APB

We know that all angles of a triangle sum up together to make 180°

So bisected A + bisected B + angle APB = 180°

=> 105° + angle APB = 180° (since we take out bisected A + bisected B = 105°)

=> angle APB = 180 - 105°

angle APB = 75°

Your answer is 75°


3) Let the angles be 3x + 5x + 7x + 9x

So we know angle sum property

3x + 5x + 7x + 9x = 360°

=> 24x = 360°

=> x = 360/24

=> x = 15°

So first angle = 3x => 3 × 15 = 45°

Second = 5x => 5 × 15 = 75°

Third = 7x => 7 × 15 = 105°

Fourth = 9x => 9 × 15 = 135°


Hope it helps dear friend ☺️✌️