Question

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

Explain with complete calculations & justifications.

Answer 1

see attachment I hope it helps......

Answer 2

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°