Question

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Answer 1

$\bf \huge \underline{ \orange{correct \: question : }}$

Given secΘ = 10/9 , calculate all other trigonometric ratios.

$\bf \huge \underline{ \orange{answer : }}$

$\sf{sec \: \theta = \frac{hyptenuse}{adjecent} = \frac{10}{9} }$

$\bf{we \: will \: find \:the \: oppsite \: side }$

$\to \tt{according \: to \: the \: attached \: diagram }$

AB² + BC² = AC²

➜ AB² = AC² - BC²

➜ AB² = 10² - 9²

➜ AB² = 100 - 81

➜ AB² = 27

∴ AB = √27

$\sf \small{all \: other \: trigonometric \: ratios \:are \: as \: follows : }$

$\to \tt{sin \theta = \frac{opposite}{hypotenuse} = \frac{ \sqrt{27} }{10} } \\ \\ \to \tt{cos \theta = \frac{adjecent}{hypotenuse} = \frac{9}{10} } \\ \\ \to \tt{tan \theta = \frac{opposite}{adjecent} = \frac{ \sqrt{27} }{9} } \\ \\ \to \tt{cot \theta = \frac{adjecent}{opposite} = \frac{9}{ \sqrt{27} } } \\ \\ \to \tt{cosec \theta = \frac{hypotenuse}{opposite} = \frac{10}{ \sqrt{27} } }$

Answer 2

$\bf\huge \underline{\orange{correct\: question:}}$

correctquestion:

Given secΘ = 10/9 , calculate all other trigonometric ratios.

$\bf\huge \underline{\orange{answer:}}$

answer:

$\begin{lgathered}\sf{sec \: \theta = \frac{hyptenuse}{adjecent} = \frac{10}{9} } \\ \\\end{lgathered}</p><p></p><p>secθ= </p><p>adjecent</p><p>hyptenuse</p><p> </p><p> = </p><p>9</p><p>10</p><p> </p><p> </p><p> </p><p> </p><p></p><p>[tex]\bf{we \: will \: find \:the \: oppsite \: side}$wewillfindtheoppsiteside

$\begin{lgathered}\to \tt{according \: to \: the \: attached \: diagram} \\\end{lgathered}$

→accordingtotheattacheddiagram

AB² + BC² = AC²

➜ AB² = AC² - BC²

➜ AB² = 10² - 9²

➜ AB² = 100 - 81

➜ AB² = 27

∴ AB = √27

$\sf\small{all \: other \: trigonometric \: ratios \:are \: as \: follows:}$allothertrigonometricratiosareasfollows:

[tex]\begin{lgathered}\to \tt{sin\theta = \frac{opposite}{hypotenuse} = \frac{\sqrt{27}}{10}} \\ \\ \to \tt{cos\theta=\frac{adjecent}{hypotenuse} =\frac{9}{10}} \\ \\ \to\tt{tan\theta = \frac{opposite}{adjecent} = \frac{\sqrt{27}}{9}} \\ \\ \to \tt{cot\theta = \frac{adjecent}{opposite} = \frac{9}{\sqrt{27}}} \\ \\ \to \tt{cosec \theta = \frac{hypotenuse}{opposite} = \frac{10}{\sqrt{27}}}\end{lgathered}

→sinθ=

hypotenuse

opposite

=

10

27

→cosθ=

hypotenuse

adjecent

=

10

9

→tanθ=

adjecent

opposite

=

9

27

→cotθ=

opposite

adjecent

=

27

9

→cosecθ=

opposite

hypotenuse

=

27

10