If two sides of a triangle are equal, the angles opposite to these sides are

not equal

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right angles

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equal

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supplementary

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Solution

The correct option is C
equal

Draw a △ABC in which AB =AC,

Draw the bisector AD of ∠A, which meets BC, in D.

Since, AB = AC, ∠BAD = ∠CAD, AD = AD

△ABD ≅ △ACD (by SAS criterion)

Then by c.p.c.t ∠B = ∠C.

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Similar questions

Q. In an isosceles triangle, if the base angles are equal, then sides opposite to these angles are equal.

Q. If two sides of a triangle are equal the angles opposite to them are also equal why?

Which of the following statements are true (T) and which are false (F):

(i) Sides opposite to equal angles of a triangle may be unequal.

(ii) Angles opposite to equal sides of a triangle are equal.

(iii) The measure of each angle of an equilateral triangle is 60°.

(iv) If the altitude from one vertex of a triangle bisects the opposite side, then the triangle may be isosceles.

(v) The bisectors of two equal angles of a triangle are equal.

(vi) If the bisector of the vertical angle of a triangle bisects the base, then the triangle may be isosceles.

(vii) The two altitudes corresponding to two equal sides of a triangle need not be equal.

(viii) If any two sides of a right triangle are respectively equal to two sides of other right triangle, then the two triangles are congruent.

(ix) Two right triangles are congruent if hypotenuse and a side of one triangle are respectively equal equal to the hypotenuse and a side of the other triangle.

Q. Prove that if two angles of a triangle are equal, then opposite sides to them are also equal.

Q. If two angles of a triangle are equal, then prove that sides opposite to them are also equal.

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If two sides of a triangle are equal, the angles opposite to these sides are

The correct option is C equal Draw a △ABC in which AB =AC, Draw the bisector AD of ∠A, which meets BC, in D. Since, AB = AC, ∠BAD = ∠CAD, AD = AD △ABD ≅ △ACD (by SAS criterion) Then by c.p.c.t ∠B = ∠C.

The correct option is C
equal

Draw a △ABC in which AB =AC,

Draw the bisector AD of ∠A, which meets BC, in D.

Since, AB = AC, ∠BAD = ∠CAD, AD = AD

△ABD ≅ △ACD (by SAS criterion)

Then by c.p.c.t ∠B = ∠C.