Advertisements
Advertisements
प्रश्न
Side BC of a triangle ABC has been produced to a point D such that ∠ACD = 120°. If ∠B = \[\frac{1}{2}\]∠A is equal to
विकल्प
80°
75°
60°
90°
Advertisements
उत्तर
In the given problem, side BC of ΔABC has been produced to a point D. Such that ∠ACD = 120° and `∠B = 1/2 ∠A` . Here, we need to find ∠A

Given `∠B = 1/2 ∠A`
We get, ∠A = 2∠B .........(1)
Now, using the property, “exterior angle of a triangle is equal to the sum of two opposite interior angles”, we get,
In ΔABC
∠ACD = ∠A + ∠B
120° = 2∠B + ∠B
120° = 3∠B
`∠B = (120°)/3`
∠B = 40°
Also, ∠A = 2∠B(Using 1)
∠ A = 2 (40°)
= 80°
Thus, ∠A = 80°
APPEARS IN
संबंधित प्रश्न
In a ΔABC, ∠ABC = ∠ACB and the bisectors of ∠ABC and ∠ACB intersect at O such that ∠BOC = 120°. Show that ∠A = ∠B = ∠C = 60°.
Is the following statement true and false :
Sum of the three angles of a triangle is 180 .
Is the following statement true and false :
All the angles of a triangle can be greater than 60°.
In Δ ABC, if u∠B = 60°, ∠C = 80° and the bisectors of angles ∠ABC and ∠ACB meet at a point O, then find the measure of ∠BOC.
Calculate the unknown marked angles of the following figure :

Calculate the angles of a triangle if they are in the ratio 4: 5: 6.
Can a triangle together have the following angles?
85°, 95° and 22°

As shown in the figure, Avinash is standing near his house. He can choose from two roads to go to school. Which way is shorter? Explain why.
The angles of the triangle are 3x – 40, x + 20 and 2x – 10 then the value of x is
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.
