If tan theta = startfraction 11 over 60 endfraction, what is the value of cot theta?

Mathematics

1 answer:

Dennis_Churaev [7]1 year ago

6 0

The value of $cot\theta$ when $tan\theta =\frac{11}{60}$ comes to be $\frac{60}{11}$ .

Given that trigonometric ratio:

$tan\theta = \frac{11}{60}$

<h3>What is the tangent of an angle?</h3>

The tangent of an angle is the ratio of the opposite side(to that angle) to the adjacent side(to that angle).

So, for the given problem

Opposite side to $\theta$ = 11

Adjacent side to $\theta$ = 60

So, $cot \theta =\frac{AdjacentSide }{OppositeSide}$

$cot\theta =\frac{60}{11}$

Therefore, the value of $cot\theta$ when $tan\theta =\frac{11}{60}$ comes to be $\frac{60}{11}$ .

To get more about trigonometric ratios visit:

brainly.com/question/24349828

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A ball is thrown vertically upwards from the ground. It rises to a height of 10m and then falls and bounces. After each bounce,

Citrus2011 [14]

${\huge{\fcolorbox{yellow}{red}{\orange{\boxed{\boxed{\boxed{\boxed{\underbrace{\overbrace{\mathfrak{\pink{\fcolorbox{green}{blue}{Answer}}}}}}}}}}}}}$

(i)

$\sf{a_n = 20 \times {( \frac{2}{3} )}^{n - 1} }$

(ii)

$\sf S_n = 60 \{1 - { \frac{2}{3}}^{n} \}$

Step-by-step explanation:

$\underline\red{\textsf{Given :-}}$

height of ball (a) = 10m

fraction of height decreases by each bounce (r) = 2/3

$\underline\pink{\textsf{Solution :-}}$

(i) We will use here geometric progression formula to find height an times

${\blue{\sf{a_n = a {r}^{n - 1} }}} \\ \sf{a_n = 20 \times { \frac{2}{3} }^{n - 1} }$

(ii) here we will use the sum formula of geometric progression for finding the total nth impact

$\orange {\sf{S_n = a \times \frac{(1 - {r}^{n} )}{1 - r} }} \\ \sf S_n = 20 \times \frac{1 - ( { \frac{2}{3} })^{n} }{1 - \frac{2}{3} } \\ \sf S_n = 20 \times \frac{1 - {( \frac{2}{3}) }^{n} }{ \frac{1}{3} } \\ \sf S_n = 3 \times 20 \times \{1 - ( { \frac{2}{3}) }^{n} \} \\ \purple{\sf S_n = 60 \{1 - { \frac{2}{3} }^{n} \}}$