मराठी

ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C. - Mathematics

Advertisements
Advertisements

प्रश्न

ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.

बेरीज
Advertisements

उत्तर

ABC is a right angled triangle in which

∠A = 90°

and AB = AC

In △ABC,

AB = AC

⇒ ∠C = ∠B     …(I)      ...[Angles opposite to equal sides]

Now, in △ABC,

∠A + ∠B + ∠C = 180°     ...[Angle Sum Property of a △)

⇒ 90° + ∠B + ∠B = 180°     ...[∵ ∠A = 90° (Given) and ∠B = ∠C from (I)]

⇒ 2∠B = 180° – 90°

⇒ 2∠B = 90°

⇒ ∠B = 45°

Also, ∠C = ∠B

⇒ ∠C = 45°

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Triangles - Exercise 7.2 [पृष्ठ १२४]

APPEARS IN

एनसीईआरटी Mathematics [English] Class 9
पाठ 7 Triangles
Exercise 7.2 | Q 7 | पृष्ठ १२४

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see the given figure). If AD is extended to intersect BC at P, show that

  1. ΔABD ≅ ΔACD
  2. ΔABP ≅ ΔACP
  3. AP bisects ∠A as well as ∠D.
  4. AP is the perpendicular bisector of BC.


Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ΔPQR (see the given figure). Show that:

  1. ΔABM ≅ ΔPQN
  2. ΔABC ≅ ΔPQR


BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.


Prove that in a quadrilateral the sum of all the sides is greater than the sum of its diagonals.


In the following figure, BA ⊥ AC, DE ⊥ DF such that BA = DE and BF = EC. Show that ∆ABC ≅ ∆DEF.


ABC and DBC are two triangles on the same base BC such that A and D lie on the opposite sides of BC, AB = AC and DB = DC. Show that AD is the perpendicular bisector of BC.


Two lines l and m intersect at the point O and P is a point on a line n passing through the point O such that P is equidistant from l and m. Prove that n is the bisector of the angle formed by l and m.


Line segment joining the mid-points M and N of parallel sides AB and DC, respectively of a trapezium ABCD is perpendicular to both the sides AB and DC. Prove that AD = BC.


ABC is a right triangle such that AB = AC and bisector of angle C intersects the side AB at D. Prove that AC + AD = BC.


ABCD is quadrilateral such that AB = AD and CB = CD. Prove that AC is the perpendicular bisector of BD.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×