Question

If sin theta 6/7 and theta is an acute angle find the other five trigonometric function values for theta

Answer 1

Given:

sin theta 6/7 and theta is an acute angle

To find:

If sin theta 6/7 and theta is an acute angle find the other five trigonometric function values for theta

Solution:

From given, we have,

sin theta 6/7 and theta is an acute angle  

we know that sin theta is a ratio of the opposite side by the hypotenuse.

So, we have, have,

h² = o² + a²

7² = 6² + a²

49 - 36 = a²

∴ a = √13

Thus the adjacent side is √13.

The trigonometric ratios are:

cos theta = a/h = √13/7

tan theta = o/a = 6/√13

sec theta = h/a = 7/√13

coses theta = h/o = 7/6

cot theta = a/o = √13/7

Therefore, the other five trigonometric function values for theta are: cos theta = √13/7, tan theta = 6/√13, sec theta = 7/√13, coses theta = 7/6 and cot theta = √13/7

Answer 2

$Given \:sin \:theta = \frac{6}{7} \: --(1)$

$i) Cos \theta =\sqrt{ 1- sin^{2} \theta }$

$= \sqrt{ 1 - \Big( \frac{6}{7}\Big)^{2}}$

$= \sqrt{ 1 - \frac{36}{49}}$

$= \sqrt{ \frac{49-36}{49}}$

$= \sqrt{ \frac{13}{49}}$

$\therefore cos \theta = \frac{\sqrt{13}}{7} \: --(2)$

$ii) tan \theta = \frac{sin \theta }{cos \theta }$

$= \frac{ \frac{6}{7}}{ \frac{\sqrt{13}}{7}}$

$\therefore tan \theta = \frac{6}{\sqrt{13}} \: ---(3)$

$iii ) cot \theta = \frac{1}{tan \theta }$

$= \frac{1}{\frac{6}{\sqrt{13}}}$

$\therefore cot \theta = \frac{\sqrt{13}}{6} \: --(4)$

$iv) Cosec \theta = \frac{1}{sin \theta }$

$= \frac{1}{\frac{6}{7}}$

$\therefore Cosec \theta = \frac{7}{6} \: --(5)$

$v) Sec \theta = \frac{1}{cos \theta }$

$= \frac{1}{ \frac{\sqrt{13}}{7}}$

$\therefore Sec \theta = \frac{7}{\sqrt{13}}\: --(6)$

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