Question

Can someone explain how tan(alpha) = f here?

Answer 1

There no rule of trig that says it’s true.

Perhaps for *this particular* right triangle, you are being given extra information beyond standard rules?

Answer 2

It isn't.

Answer 3

Are you winding us up, OP? :)

Answer 4

Assume this is a right triangle, and that the hypotenuse has length *f*. Let *a* be the opposite sidelength, and *b* the adjacent. We have that sinα = *a*/*f*, and cosα = *b*/*f*. We further have that tanα = *a*/*b* = (*f* sinα)/(*f* cosα). That is, the only way tanα = *f*, is when sinα/cosα = *f*, or equivalently sinα = *f* cosα.     
-1 <= sinα <= 1, and same holds for cosine. This forces your *f* = 1, and your angle to α = 45°.

Answer 5

I don't see why that should be true

Use SOHCAHTOA

Answer 6

Is this a right triangle even?

Answer 7

If tan(alpha)=f, and f²=y²+x², and tan(alpha)=y/x then y²+x²=tan²(alpha)=y²/x².  
y²x²+x⁴=y²  
y²(x²-1)=-x⁴  
y²=x⁴/(-x²+1)  
Assuming y and x are real-valued and positive, this means x² must be between 0 and 1 (and therefore x is,  as well), but there are no particular constraints, beyond that. y is a function of x, and it's one-to-one for 0<x<1, so as your triangle gets taller, the x side must get closer and closer to 1, asymptotically.

Answer 8

Tan(a) = opposite/adjacent

Answer 9

I’ll just add that tan alpha = f doesn’t make sense physically since f has units of length. But sure, if length is just a number you could rescale any right angles triangle, keeping alpha the same, until the hypotenuse is tan alpha or cos alpha or any positive number you want.

Answer 10

Suppose the adyacent catetus is a and opposite catetus is b. Then, by definition of tan(alpha), we have f=a/b (*) and, by Pythagoras, we have also a² + b² = f² (**). Then wey have have a equation's system which incognites ate a and b.

f=a/b ( *)

a²+b²=f² (**)


By ( *) wey have a=f•b (***)


Then, reemplacing ( ***) in (**) and solving for b we have


(f•b)² + b² = f²

f² • b²  + b² = f²

(f² + 1) • b² = f²

b² = f² / (f² + 1)

b = (f² / (f² + 1))^(1/2)

b = f /((f² + 1))^(1/2)

b = f • (f² + 1)^(1/2) / (f² + 1). (****)

Then, reemplacing (****) in (***) a solving for a we have

a=b•f

a = (f • (f² + 1)^(1/2) / (f² + 1)) • f

a = f² • (f² + 1)^(1/2) / (f² + 1)