Question

now how bout this?

Answer 1

k?

Answer 2

Confused much?

Answer 3

no

Answer 4

i just like asking questions

Answer 5

I see.

Answer 6

to ask and answer i use google

Answer 7

In \(△ADB,∠ADB=(180−18−30)^∘=132^∘\)

Applying sine law in \(△ADB\),

\(\Large\frac{AB}{sin132^∘}=\frac{AD}{sin30^∘}⟹AD=\frac{AB}{2sin48^∘}\)

as \(sin132^∘=sin(180−132)^∘=sin48^∘\)

\(∠ABC=∠BAC=\Large\frac{1806^∘−96^∘}{2}=42^∘\)

Applying sine law in \(△ABC\),

\(\Large\frac{AC}{sin42^∘}=\frac{AB}{sin96^∘}⟹\frac{AC}{AB}=\frac{sin42^∘}{sin96^∘}=\frac{cos48^∘}{2sin48^∘cos48^∘}\)
(applying \(sin~2A=2~sin~A~cos~A\))

So,
\(AC=\Large\frac{AB}{2sin48^∘}⟹AC=AD\)

So, \(∠ACD=∠ADC=\Large\frac{(180−24)^∘}{2}=78^∘\)

Answer 8

AWESOME YOU ARE SOOO COOL

Answer 9

That one took forever

Answer 10

sorry but that was awesome