Advertisements
Advertisements
प्रश्न
One of the base angles of an isosceles triangle is 52°. Find its angle of the vertex.
Advertisements
उत्तर
Each of the base angles of an isosceles
Δ ABC = 52°
∴ ∠B = ∠C = 52°
But ∠A + ∠B + ∠C = 180°..........(Angles of a triangle)
⇒ ∠A + 52° + 52° = 180°
⇒ ∠A + 104° = 180°
⇒ ∠A = 180°− 104° = 76°
Hence ∠A = 76°
APPEARS IN
संबंधित प्रश्न
In the given figure, AM ⊥ BC and AN is the bisector of ∠A. If ∠B = 65° and ∠C = 33°, find ∠MAN.
In the given figure, AE bisects ∠CAD and ∠B= ∠C. Prove that AE || BC.
Define a triangle.
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.
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
In ΔRST (See figure), what is the value of x?
Calculate the unknown marked angles of the following figure :
Find the unknown marked angles in the given figure:
If an angle of a triangle is equal to the sum of the other two angles, find the type of the triangle
In the following figure, PQ ⊥ RQ, PQ = 5 cm and QR = 5 cm. Then ∆PQR is ______.
