Question

plz
solve this problem (╥﹏╥)(╥﹏╥)​

Answer 1

Answer:

Plz refer to the attachment

Hope its helps u..

Answer 2

Answer:

0

Step-by-step explanation:

In the figure, given tanA=√3

tan A= BC/AB

√3/1=BC/AB

so, BC= √3 and AB = 1

therefore, AC²=AB²+BC²

AC²= 1²+(√3)²

AC= 2

so, cosA= AB/AC= 1/2

III ly, cosC=BC/AC=√3/2

sinA=√3/2

sinC= 1/2

cosA. cosC-sinA. sinC=1/2.√3/2-(√3/2.1/2)

=1/2.√3/2-√3/2.1/2

= 0