Question

вычислите tga

Answer 1

sinα=5/13         π/2<α<π (вторая четверть)  ⇒   tgα<0     tg=?

cosα=√(1-sin²α)=√(1-(25/169))=√(144/169)=+/-12/13     ⇒

cosα=-12/13.

tgα=sinα/cosα=(5/13)/(12/13)=5/12.

ответ: tgα=5/12.

Answer 2

It seems there might be a misunderstanding or a typo in your question. "tga" does not represent any common mathematical operation or notation that I am aware of. If you meant something else or if there is a specific context to the term "tga," please provide more information or clarify your question, and I'll be happy to help you further.