Please solve question no.18.

Dear Student
GIVEN :-

abcd IS A RHOMBUS
so,
ab=bc=cd=ad

ac = 8 cm

bd = 6 cm

Now,

we know that In rhombus diagonal bisect each other at center and angle formed is  90°.


oa= oc = (ac/2) = 4cm

ob = od = (bd/2) =3cm

Now,

In right △ boc
ob{}^{2} + oc {}^{2} = bc {}^{2} \\ \\( 3cm) {}^{2} + (4cm) {}^{2} = bc{}^{2} \\ \\ 9cm {}^{2} + 16cm {}^{2} = bc {}^{2} \\ \\ 25cm {}^{2} = bc {}^{2} \\ \\ \sqrt{25cm {}^{2} } = bc \\ \\ 5cm = bc

Now in triangle ocd

sin space angle o c d space equals fraction numerator o d over denominator c d end fraction equals 3 over 5
Regards​​​​​​

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