Question

How do you do this geometry question?

Answer 1

Answer: B: 150°

Step-by-step explanation:

$|EF|=\sqrt{1^2-(\dfrac{1}{2})^2}}=\sqrt{\dfrac{3}{4} } =\dfrac{\sqrt{3} }{2} \\\\tan(\alpha)=\dfrac{1-\dfrac{\sqrt{3} }{2} }{\dfrac{1}{2} } =2-\sqrt{3} \Longrightarrow\ \alpha=15^o \\\\mes\widehat{BEC}=180^o-2*15^o=150^o\\\\Answer\ B$

Answer 2

Answer:

Ans: A 160

Step-by-step explanation:

(Hope this helps have a great day :)