Question

ABCD is a rhombus in which angle A=60°. Find the ratio AC:BD
[Figure attached].

Answer 1

Hey there!
We will use the following things in our solution :
#1 : All sides of a rhombus are equal
#2 : The diagonals bisect each other and are mutual perpendicular bisectors.
#3 : Relation between sides in a equilateral triangle and Right angled triangle.

First of all, A0 = AC/2 [ Diagonals bisect each other ]

Now, let's Consider that AD = AB [ A sides of a rhombus are equal ]

Also, One of the angle is given.
∠A = 60°

Now, Consider the triangle ΔABD,
We have AB = AD,
So It is an isosceles triangle.
So ∠B = ∠D

Now Use Angle Sum property;
∠A + ∠B + ∠C = 180
60 + 2∠B = 180
∠B = 120/2 = 60°

So , We found that,
∠A = ∠B = ∠C.

So, It is equilateral triangle.

From this , AB = AD = BD. [ Equation A ]

In ΔAOD,
∠AOD = 90° [ Diagonals are mutually perpendicular bisector ]

ΔAOD is right angled triangle.
So,
AD² = AO² + OD² [ Pythagoras theorem ]

OD = BD/2 = AD/2 [ AD = BD ]

AD² = ( AD/2)² + AO²
AD² - AD²/4 = AO²
3AD²/4 = AO²

AO = √3AD/2
AC = 2 * AO = 2 ( √3/2 AD) = √3 AD

From Equation A, BD = AD

Now, AC : BD = √3 AD : AD = √3 : 1

Answer 2

Answer:

Rhombus ABCD, => diagonals AC & BD bisect at right angles. Point of intersection is O.

& also diagonals bisect interior angles of it. So, if in right triangle AOB

tan 30° = OB/OA = 1/√3

=> OB = x & OA = √3x

=> DB = 2x

& AC = 2√3x

=> DB/AC = 2x/2√3x

=> DB/AC = 1/√3

=> Ratio of diagonal opposite to smaller 60° angle to that of diagonal opposite to greater 120° angle = 1 : √3

I Hope It Will Help!

^_^