Question

In a rhombus ABCD, the altitude from D to side AB bisects AB. Find the angles of the rhombus.

Answer 1

Given:

ABCD is a rhombus.

The altitude from D to AB meets AB at its midpoint i.e., the perpendicular from D to AB bisects AB.

Step-wise calculation:

1. Let the foot of the altitude be M. 

So, M is the midpoint of AB and DM ⟂ AB.

2. Put coordinates:

Let A = (–1, 0), B = (1, 0). 

So, AB = 2 and M = (0, 0).

Since DM ⟂ AB and passes through M, let D = (0, h) for some h > 0.

3. All sides of a rhombus are equal.

So, AD = AB = 2.

Compute AD:

AD = Distance between A(–1, 0) and D(0, h)

= `sqrt((0 + 1)^2 + (h - 0)^2`

= `sqrt(1 + h^2)`

Thus, `sqrt(1 + h^2) = 2`

⇒ 1 + h² = 4

⇒ h² = 3 

⇒ `h = sqrt(3)`

4. Now find angle A.

Vectors from A: 

AB = (2, 0)

`AD = (1, sqrt(3))`

Cosine of `∠A = (AB xx AD)/(|AB||AD|)` 

= `(2 xx 1 + 0 xx sqrt(3))/(2 xx 2)` 

= `2/4`

= `1/2` 

Hence, ∠A = 60°.

5. Opposite and adjacent angles in a rhombus are equal/supplementary.

So, the four angles are 60°, 120°, 60°, 120°.

The angles of rhombus ABCD are 60°, 120°, 60° and 120°.