Question

ABCD is a rhombus in which the altitude from D to side AB bisects AB. Find the angles of the rhombus

Answer 1

let @ is the angle of DAB in rohmbus 

according to Rohmbus property , 
all side of rohmbus are equal .
e.g 
AB = BC =CD =DA 

now , Draw a rough diagram of Rohmbus ABCD 

altitude from D meet in AB at M 
a/c to question ,

AM = MB =1/2 AB = 1/2BC =1/2 CD =1/2AD 

here ∆ ADM is right angle traingle .
where 
AD is hypotenuse
DM is altitude 
and AM is base 

formula of cos@ =base/hypotenuse 

so,
cos@ = AM/AD 

from above 

cos@ = 1/2AD /AD = 1/2 
but we know cos60° =1/2

so, cos@ = cos 60°

@= 60° =angle DAB = angle BCD 

we also know rohmbus is ||gm
so, 
angle ADC +angle DAB =180°

angle ADC = 180° - 60° =120°

angle ADC = angle ABC = 120°

so, 
angles of rohmbus are

60° , 120° , 60° ,120°

Hope This Helps :)