Advertisements
Advertisements
Question
In a triangle ABC, BC = AC and ∠ A = 35°. Which is the smallest side of the triangle?
Advertisements
Solution
In ΔABC,
BC = AC ...(given)
⇒ ∠A = ∠B = 35°
Let ∠C = x°
In ΔABC,
∠A + ∠B + ∠C = 180°
35° + 35° + x = 180°
70° + x° = 180°
x° = 180° - 70°
x° = 110°
∠C = x° = 110°
Hence, ∠A = ∠B = 35° and ∠C = 110°
In ΔABC, the greatest angle is ∠C.
As the smallest angles are ∠A and ∠B,
smallest sides are BC and AC.
RELATED QUESTIONS
Show that in a right angled triangle, the hypotenuse is the longest side.
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see the given figure). Show that ∠A > ∠C and ∠B > ∠D.

In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.
Name the greatest and the smallest sides in the following triangles:
ΔDEF, ∠D = 32°, ∠E = 56° and ∠F = 92°.
Name the smallest angle in each of these triangles:
In ΔPQR, PQ = 8.3cm, QR = 5.4cm and PR = 7.2cm
Name the smallest angle in each of these triangles:
In ΔXYZ, XY = 6.2cm, XY = 6.8cm and YZ = 5cm
In ΔABC, the exterior ∠PBC > exterior ∠QCB. Prove that AB > AC.
D is a point on the side of the BC of ΔABC. Prove that the perimeter of ΔABC is greater than twice of AD.
In ΔPQR, PS ⊥ QR ; prove that: PQ > QS and PQ > PS
In ΔPQR, PS ⊥ QR ; prove that: PQ > QS and PR > PS
