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Teaching Since:	May 2017
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MBA (IT), PHD
Kaplan University
Apr-2009 - Mar-2014

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Professor
University of Santo Tomas
Aug-2006 - Present

Category > Geometry Posted 30 Oct 2017 My Price 11.00

If sin theta = 3/5 and theta is acute, how do you calculate cos theta and tan theta?

$\cos θ = \frac{4}{5}, and, \tan θ = \frac{3}{4}$

We will use the Identity $: \sin^{2} θ + \cos^{2} θ = 1$

$∴ \cos^{2} θ = 1 - \sin^{2} θ = 1 - {(\frac{3}{5})}^{2} = 1 - \frac{9}{25} = \frac{16}{25}$

$∴ \cos θ = \sqrt{\frac{16}{25}} = \pm \frac{4}{5}$

But we know that theta is acute , so, it is in the first quadrant, in which, all the Trgo. ratios are $+ v e$

Hence, $\cos θ = + \frac{4}{5}$

Finally, $\tan θ = \frac{\sin θ}{\cos θ}$

..............[Defn.]

$= \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}$

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Dr Nick

(3)

Status NEW Posted 30 Oct 2017 01:10 PM My Price 11.00

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