This kind of exercise can be real drudgery. But it's almost all simple arithmetic, so better you than me. I'll do one of these for you, which will show you how to do the other one.
a). sin · cot / sec
You're supposed to know that [ cotangent = cosine/sine ]
and [ secant = 1/cosine ].
Then the problem becomes
sin · (cos/sin) / (1/cos) = cos²
Aw shucks, I might as well also set up 'b)' for you:
b). cos · csc / tan
You're supposed to know that [ cosecant = 1/sine ]
and [ tangent = sine/cosine ].
Then the problem becomes
cos · (1/sin) / (sin/cos) = (cos/sin)²
Now simply plug in the given values of cos A and sin A .
And may I compliment you on your nail care !
Answer:
y would be 80...because alternate exterior angles are equivalent
Step-by-step explanation:
in the picture is that what you have or no?
