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tia_tia [17]
3 years ago
15

Help mee I need this

Geography
2 answers:
Mademuasel [1]3 years ago
5 0

XD XD XD XD XD XD XD XD XD XD XD XD

galben [10]3 years ago
3 0
I’m not sure of answer but you can ask the same question in internet and it will give you a good image
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Suppose 0 is an angle in the standard position whose terminal side is in Quadrant IV and cot0= -6/7 Find the exact values of the
Sergio039 [100]
<h2>\sin \theta=\dfrac{p}{h} =\dfrac{-7}{\sqrt{85} }, \cos \theta=\dfrac{b}{h} =\dfrac{6}{\sqrt{85}}, \tan \theta=\dfrac{p}{b} =\dfrac{-7}{6},</h2><h2>\sec \theta=\dfrac{h}{b} =\dfrac{\sqrt{85} }{6} and \csc \theta=\dfrac{h}{p} =\dfrac{-\sqrt{85}}{ 7}</h2>

Explanation:

Given,

\cot \theta= \dfrac{-6}{7}

To find, the exact values of the five remaining trigonometric functions of \theta = ?

We know that,

\cot \theta= \dfrac{-6}{7}=\dfrac{b}{p}

Where, b = base and p = perpendicular

By Pythagoras's theorem,

Hypotaneous, h=\sqrt{p^2+b^2}

=\sqrt{(-6)^2+7^2}=\sqrt{36+49} =\sqrt{85}

In IVth quadrant,

\cot \theta and \sec \theta are positive.

\sin \theta=\dfrac{p}{h} =\dfrac{-7}{\sqrt{85} }, \cos \theta=\dfrac{b}{h} =\dfrac{6}{\sqrt{85}}, \tan \theta=\dfrac{p}{b} =\dfrac{-7}{6},

\sec \theta=\dfrac{h}{b} =\dfrac{\sqrt{85} }{6} and \csc \theta=\dfrac{h}{p} =\dfrac{-\sqrt{85}}{ 7}

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