Advertisements
Advertisements
प्रश्न
Find, giving a reason, the unknown marked angles, in a triangle drawn below:

Advertisements
उत्तर
We know that,
Exterior angle of a triangle is always equal to the sum of its two interior opposite angles (property)
x+115° = 180°
(linear property of angles)
⇒ x = 180°- 115° ⇒ x = 65°
∴115° = x + y
⇒ 115° = 65° + y
⇒ y= 115° – 65° =50°
y = 50°
APPEARS IN
संबंधित प्रश्न
Is the following statement true and false :
All the angles of a triangle can be equal to 60°.
Is the following statement true and false :
An exterior angle of a triangle is greater than the opposite interior angles.
In a Δ ABC, the internal bisectors of ∠B and ∠C meet at P and the external bisectors of ∠B and ∠C meet at Q, Prove that ∠BPC + ∠BQC = 180°.
In Δ ABC, BD⊥ AC and CE ⊥ AB. If BD and CE intersect at O, prove that ∠BOC = 180° − A.
If one angle of a triangle is equal to the sum of the other two angles, then the triangle is
Find the value of the angle in the given figure:

Find the unknown marked angles in the given figure:

Find x, if the angles of a triangle is:
x°, x°, x°
One angle of a right-angled triangle is 70°. Find the other acute angle.
The angles of the triangle are 3x – 40, x + 20 and 2x – 10 then the value of x is
