Green Yoda
Hi Φ
- Joined
- Mar 28, 2015
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- HSC
- 2017
Can you elaborate on how DC = 2sin30
Using right-angled trig. in triangle DCA (which, by the way, is just the standard "30, 60, 90" triangle).Can you elaborate on how DC = 2sin30
Yes that is correct.Yup got it, So what i did was first find AC by Cos 30 = AC/2 which is x = cos 30 x 2 = √ 3/2 x 2, therefore AC= √ 3 .
Now I found DC by sin 30 = DC/2, there fore DC = sin 30 x 2 which is 1/2 x so therefore DC = 1
As angles D and C are same so that means DC and BC are same so therefore BC = 1.
Now to find BD, AC-BC which is √ 3 -1.
Is this the correct method?