Advertisements
Advertisements
प्रश्न
Find, giving a reason, the unknown marked angles, in a triangle drawn below:
उत्तर
We know that,
Exterior angle of a triangle is always equal to the sum of its two interior opposite angles (property)
110° = 2x + 3x
5x – 110°
x = `110^circ/5`
x = 22°
∴ 2x = 2 x 22 = 44°
3x = 3 x 22 = 66°
APPEARS IN
संबंधित प्रश्न
ABC is a triangle in which ∠A — 72°, the internal bisectors of angles B and C meet in O.
Find the magnitude of ∠BOC.
Fill in the blank to make the following statement true:
Sum of the angles of a triangle is ....
In ΔABC, ∠B = ∠C and ray AX bisects the exterior angle ∠DAC. If ∠DAX = 70°, then ∠ACB =
An exterior angle of a triangle is 108° and its interior opposite angles are in the ratio 4 : 5. The angles of the triangle are
The bisects of exterior angle at B and C of ΔABC meet at O. If ∠A = x°, then ∠BOC =
Classify the following triangle according to sides:
Can you draw a triangle with 25°, 65° and 80° as angles?
One of the angles of a triangle is 65°. If the difference of the other two angles is 45°, then the two angles are
If an angle of a triangle is equal to the sum of the other two angles, find the type of the triangle
Can we have two acute angles whose sum is a right angle? Why or why not?