If cos=3/5, then tan=

2 answers:

6 0

This is a 3-4-5 triangle.

Remember,
SOH-CAH-TOA

S(ine) = O(pposite)/H(ypotenuse)
C(osine) = A(djacent)/H(ypotenuse)
T(angent) = O(pposite)/A(djacent)

In Cosine, the adjacent = 3, and the Hypotenuse = 5

Hypotenuse is the longest side.

3,4,5 is the triangle that fits inside, therefore, the Opposite is 4, and Adjacent is 3

T(angent) = O(pposite)/A(djacent)

Plug in the numbers

T = 4/3

tan = 4/3 is your answer

hope this helps

Kazeer [188]3 years ago

5 0

So cosФ = 3/5 means that ratio of length of adjacent side to length of hypotenuse side is 3:5

So then we can assume adjacent side is 3 and hypotenuse is 5.

By Pythagoras<span>' </span>Theorem, we have √(5² - 3²) = 4, so then length of opposite side would be 4.

Now tanФ is ratio of length of opposite side to length of adjacent side, which would be 4/3.

Final answer: 4/3

Hope this helps.

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7 0

4 years ago

Question 2 of 5

AlekseyPX

Given:

The different recursive formulae.

To find:

The explicit formulae for the given recursive formulae.

Solution:

The recursive formula of an arithmetic sequence is $f(n)=f(n-1)+d, f(1)=a,n\geq 2$ and the explicit formula is $f(n)=a+(n-1)d$ , where a is the first term and d is the common difference.

The recursive formula of a geometric sequence is $f(n)=rf(n-1), f(1)=a,n\geq 2$ and the explicit formula is $f(n)=ar^{n-1}$ , where a is the first term and r is the common ratio.

The first recursive formula is:

$f(1)=5$

$f(n)=f(n-1)+5$ for $n\geq 2$ .

It is the recursive formula of an arithmetic sequence with first term 5 and common difference 5. So, the explicit formula for this recursive formula is:

$f(n)=5+(n-1)(5)$

$f(n)=5+5(n-1)$

Therefore, the correct option is A, i.e., $f(n)=5+5(n-1)$ .

The second recursive formula is:

$f(1)=5$

$f(n)=3f(n-1)$ for $n\geq 2$ .

It is the recursive formula of a geometric sequence with first term 5 and common ratio 3. So, the explicit formula for this recursive formula is:

$f(n)=5(3)^{n-1}$

Therefore, the correct option is F, i.e., $f(n)=5(3)^{n-1}$ .

The third recursive formula is:

$f(1)=5$

$f(n)=f(n-1)+3$ for $n\geq 2$ .

It is the recursive formula of an arithmetic sequence with first term 5 and common difference 3. So, the explicit formula for this recursive formula is:

$f(n)=5+(n-1)(3)$