+13 Can someone please help me understand how to break this down, this was no where in my homework or lecture but on the test twice?
Linking screenshots of the quetsions
Hello,
The first question:
So, the first question wants us to find a side "x" with a given angle & a side length
and It also gives us that:
x=7cot(39)
so 39 is the angle in (degrees)
well you might be questioning what is 7??
Let's first know cot
cot is co-tangent
We also know that cotangent= Adj/opposite
While tangent=Opposite/Adj
Let's first find cot(39)
cot(39)= Adj/opp = x/7
Well we want to solve for x
Since we know that Cot(39)=x/7
Therefore let's multiply both sides by 7
we conclude that:
7*Cot(39)=x
So let's look at triangles..
The correct triangle have the following just from the information we got:
1-the opposite side of the angle is =7
2-The adjacent is x
F. is the correct :)
you chose "E." You simply found tangent.
because tangent = opp/adj while cotangent=adj/opp
Second question:
Again we have:
4 cos 29
where 29 is the angle in (degrees)
and 4 is a side (We will know soon)
Well first let's know cosine
cos 29=adj/hyp
so let's find the adj side to cos 29
Well it must be "x" because cos(29)=x/4
so multiplying both sides by 4 we get that 4*cos(29)=x
So option E. Is correct. Also notice that X and 4 can't be the opposite side of the angle because cosine is adj/hyp
Hope that helped :) and good luck!
+13 I tried applying that to another question, and I for the life of me can't understand why the hypotenuse is x on this one.
I'll show you my notes and the question, hopefully you're still around to explain what I did wrong.
Sorry.
+36916 sec = 1/cos = 1/(adj/hyp) = hyp/adj
From your Q x=3 hyp/adj of the angle x = 3 x/3 of the 42 angle
Just a pattern I noticed:
x=3 sec 42
let it be in that form:
x= r sec z (angle)
sec z= r/f
where f is the adjacent side. (Changes depends on the trigonometric function etc)
while r is the hypotenus
so r=hyp.
in this case:
x= 3 sec 42
r=3
so x is the adjacent according to the trigonometric function sec hyp/adj
+13 Thx, I got it now, aced my test also :D
Pattern I went with made it so easy for me,
I can just take x=7cot(39)
divide the 7 out right away so that x over something = cot(theta)
x/7 = cot39
Then just set trig function what side it represents, in this case adj/op
x/7 = a/o
and it matches up every time x=a and 7=o adjusted to theta
