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3 years ago

13

A ramp is placed from a ditch to a main road 2ft. above the ditch. If the length of the ramp is 12ft, how far away is the bottom

of the ramp from the road?

1 answer:

Nadusha1986 [10]3 years ago

6 0

See attached picture for solution:

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2 years ago

Given that Cosecant (t) = negative StartFraction 13 Over 5 EndFraction for Pi less-than t less-than StartFraction 3 pi Over 2 En

Sergio [31]

Answer:

$\bold{cot(t) =\dfrac{12}{5}}$

Step-by-step explanation:

Given that:

$Cosec (t) = -\frac{13}5$

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That means, angle $t$ is in the 3rd quadrant.

To find:

Value of cot(t)

Solution:

First of all, let us recall what trigonometric ratios are positive and what trigonometric ratios are negative in 3rd quadrant.

In 3rd quadrant, tangent and cotangent are positive.

All other trigonometric ratios are negative.

Let us have a look at the following identity:

$cosec^2\theta -cot^2\theta =1$

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So, $cosec^2t-cot^2t=1$

$\Rightarrow (-\dfrac{13}{5})^2-cot^2t=1\\\Rightarrow (\dfrac{169}{25})-cot^2t=1\\\Rightarrow \dfrac{169}{25}-1=cot^2t\\\Rightarrow \dfrac{169-25}{25}=cot^2t\\\Rightarrow \dfrac{144}{25}=cot^2t\\\Rightarrow cot(t)=\pm\sqrt{\dfrac{144}{25}}\\\Rightarrow cot(t)=\pm\dfrac{12}{5}$

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$\bold{cot(t) =\dfrac{12}{5}}$

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3 years ago

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i think this is the ans but im not very sure

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2 years ago