In a rhombus, diagonals bisect each other at 90 degrees and also bisect the angles of the rhombus.

Given that ∠ACB = 50°, we know that the diagonal AC bisects ∠A and ∠C of the rhombus. Therefore, ∠BAC = ∠BCA = 50°.

Now, we know that in a triangle, the sum of all angles is 180°. So, in triangle ABC, ∠ABC = 180° - ∠BAC - ∠BCA = 180° - 50° - 50° = 80°.

Since ∠ABC = 80° and the diagonals bisect the angles, ∠ABD = ∠ABC = 80°.

So, ∠ADB = 80°.

Question

In a rhombus, diagonals bisect each other at 90 degrees and also bisect the angles of the rhombus.

Given that ∠ACB = 50°, we know that the diagonal AC bisects ∠A and ∠C of the rhombus. Therefore, ∠BAC = ∠BCA = 50°.

Now, we know that in a triangle, the sum of all angles is 180°. So, in triangle ABC, ∠ABC = 180° - ∠BAC - ∠BCA = 180° - 50° - 50° = 80°.

Since ∠ABC = 80° and the diagonals bisect the angles, ∠ABD = ∠ABC = 80°.

So, ∠ADB = 80°.

Knowee AI · Accepted Answer

In a rhombus, diagonals bisect each other at 90 degrees and also bisect the angles of the rhombus.

Given that ∠ACB = 50°, we know that the diagonal AC bisects ∠A and ∠C of the rhombus. Therefore, ∠BAC = ∠BCA = 50°.

Now, we know that in a triangle, the sum of all angles is 180°. So, in triangle ABC, ∠ABC = 180° - ∠BAC - ∠BCA = 180° - 50° - 50° = 80°.

Since ∠ABC = 80° and the diagonals bisect the angles, ∠ABD = ∠ABC = 80°.

So, ∠ADB = 80°.

ABCD is a rhombus such that ∠ACB = 50°. Then find ∠ADB