As shown in the figure, in triangle ABC, ∠ ACB = 60 ° AC ＞ BC, there are triangle ABC ', triangle BCA', triangle cab ' Better be more detailed

As shown in the figure, in triangle ABC, ∠ ACB = 60 ° AC ＞ BC, there are triangle ABC ', triangle BCA', triangle cab ' Better be more detailed

Er, your question is incomplete. What's the solution

If we know the height (AD = 3.1) on the BC side of ∠ BAC = 60 degree in ABC, if tanb = √ 3 / 2, we can find the area of ABC
2. If ABC is an acute triangle, find the minimum area of ABC
(1) Answer (7 / 2) √ 3
(2) If a = 0, it is 3 √ 3
It's really the right number. The question is in hand

I have the wrong number
Because tanb √ 3. So BD = 2 √ 3. AB = √ 21
SINB = √ 21 / 7. CoSb = 2 / 7 √ 3. So sinc = 3 / 14 √ 21. So BC = 7 / 3 √ 3
So s = 1 / 2 * AB * BC * SINB = 7 / 2 √ 3

As shown in the figure, AB and AC are isosceles of △ ABC, ad plane ∠ BAC, and △ BCD are isosceles triangle? Why?

Delta BCD is an isosceles triangle
Proof: ad bisects ∠ BAC
So: ∠ bad = ∠ CAD
And ab = AC, ad = ad
So: △ abd ≌ △ ACD
So: BD = CD
So, △ BCD is an isosceles triangle