Advertisements
Advertisements
प्रश्न
One angle of a triangle is 61° and the other two angles are in the ratio `1 1/2: 1 1/3`. Find these angles.
Advertisements
उत्तर
In Δ ABC,
Let ∠A = 61°
But ∠A + ∠B + ∠C = 180° ........(Angles of a triangle)
⇒ 61° + ∠B + ∠C = 180°
⇒ ∠B + ∠C = 180°− 61° = 119°
But ∠B: ∠C =`1 1/2:1 1/3=3/2:4/3`
`=(9:8)/6`
=9: 8
Let ∠B = 9x and ∠C = 8x,
then, 9x + 8x = 119°
⇒ 17x = 119°
⇒ x =`(119°)/17=7°`
∴ ∠B = 9x = 9 × 7° = 63°
∠C = 8x = 8 × 7° = 56°
APPEARS IN
संबंधित प्रश्न
In a ΔABC, if ∠A = 55°, ∠B = 40°, find ∠C.
Is the following statement true and false :
A triangle can have two right angles.
Is the following statement true and false :
All the angles of a triangle can be greater than 60°.
Fill in the blank to make the following statement true:
A triangles cannot have more than ......obtuse angles.
Calculate the unknown marked angles of the following figure :
In the given figure, express a in terms of b.
Classify the following triangle according to angle:
One of the angles of a triangle is 65°. If the difference of the other two angles is 45°, then the two angles are
Bisectors of the angles B and C of an isosceles triangle with AB = AC intersect each other at O. BO is produced to a point M. Prove that ∠MOC = ∠ABC.
Can we have two acute angles whose sum is a reflex angle? Why or why not?
