Olympiad Test: Lines and Angles

∵ In ∆ABC,
AB = AC
∴ ∠B = ∠C    ...(1)
| Angles opposite to equal sides of a triangle are equal
In ∆ABC,
∠A + ∠B + ∠C = 180°
| Sum of all the angles of a triangle is 180°
⇒ 90° + ∠B + ∠C = 180°
| ∵ ∠A = 90° (given)
⇒    ∠B + ∠C = 90°    ...(2)
From (1) and (2), we get
∠B = ∠C = 45°.

Supplementary angles sums up to 180° x + 105° = 180° x=180° - 105° = 75°

Complementary angle sums up to 90°. So x+ 45° = 90°
x = 90° - 45° = 45°

Let the angle be x°
then,complement°=(90-x)°

A/Q,
x=(90-41)
x=90-41
x=49

Let one of the two equal complementary angles be x.

Thus, 45^o is equal to its complement. 

Supplementary angles sums up to 180° since 90° + 90° = 180°, 90° is supplementary to itself

120 and x are corresponding angles and corresponding angles are equal so x = 120^°

The angles are co-interior so x+60 = 180
x=120^°

Given,
∠A = 90° and AB = AC
A/q,
AB = AC 
⇒ ∠B = ∠C (Angles opposite to the equal sides are equal.)
Now,
∠A + ∠B + ∠C = 180° (Sum of the interior angles of the triangle.)
⇒ 90° + 2∠B = 180°
⇒ 2∠B = 90° 
⇒ ∠B = 45°
Thus, ∠B = ∠C = 45°

This will be the equation x=5(90-x)
solution= x=5(90-x)              
                x+5x=90
                6x=90
                 6x/6=90/6   {divide by 6 to both the side to x}
                  x=15

ans= The angle is 15 degree and its compliment angle is 75 degree