Question

If tan thita =7/24 then find the value of sin thita and sec thita​

Answer 1

Answer:

$\displaystyle{{\boxed{\red{\pink{\sf\:\sin\:\theta\:=\:\dfrac{7}{25}}}}}}$

$\displaystyle{\boxed{\blue{\sf\:\sec\:\theta\:=\:\dfrac{25}{24}}}}$

Step-by-step-explanation:

We have given that,

$\displaystyle{\sf\:\tan\:\theta\:=\:\dfrac{7}{24}}$

We have to find the value of $\displaystyle{\sf\:\sin\:\theta\:\&\:\sec\:\theta}$

Now, we know that,

$\displaystyle{\pink{\sf\:\tan\:\theta\:=\:\dfrac{Opposite\:side}{Adjacent\:side}}}$

$\displaystyle{\implies\sf\:\dfrac{7}{24}\:=\:\dfrac{Opposite\ side}{Adjacent\ side}}$

∴ Opposite side = 7 units

Adjacent side = 24 units

Now, we know that,

$\displaystyle{\sf\:(\:Hypotenuse\:)^2\:=\:(\:Opposite\ side\:)^2\:+\:(\:Adjacent\:side\:)^2}$

$\displaystyle{\implies\sf\:H^2\:=\:(\:7\:)^2\:+\:(\:24\:)^2}$

$\displaystyle{\implies\sf\:H^2\:=\:49\:+\:576}$

$\displaystyle{\implies\sf\:H^2\:=\:625}$

$\displaystyle{\implies\sf\:H\:=\:\sqrt{625}}$

$\displaystyle{\implies\boxed{\red{\sf\:Hypotenuse\:=\:25\:units}}}$

Now, we know that,

$\displaystyle{\pink{\sf\:\sin\:\theta\:=\:\dfrac{Opposite\ side}{Hypotenuse}}}$

$\displaystyle{\implies\sf\:\sin\:\theta\:=\:\dfrac{7}{25}}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\pink{\sf\:\sin\:\theta\:=\:\dfrac{7}{25}}}}}}$

Now, we know that,

$\displaystyle{\blue{\sf\:\sec\:\theta\:=\:\dfrac{Hypotenuse}{Adjacent\ side}}}$

$\displaystyle{\implies\sf\:\sec\:\theta\:=\:\dfrac{25}{24}}$

$\displaystyle{\therefore\:\underline{\boxed{\blue{\sf\:\sec\:\theta\:=\:\dfrac{25}{24}}}}}$

Answer 2

Given : $\sf{\tan \theta = \dfrac{7}{24}}$

Need To Find : The value of $\sin \theta \:and\:\sec\theta$

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❍ Basic Formulas of Trigonometry is given by :

$\boxed { \begin{array}{c c} \\ \dag \qquad \large {\underline {\bf{ Some \:Basic\:Formulas \:For\:Trigonometry \::}}}\\\\ \sf{ In \:a \:Right \:Angled \: Triangle-:} \\\\ \sf {\star Sin \theta = \dfrac{Perpendicular}{Hypotenuse}} \\\\ \sf{ \star \cos \theta = \dfrac{ Base }{Hypotenuse}}\\\\ \sf{\star \tan \theta = \dfrac{Perpendicular}{Base}}\\\\ \sf{\star \cosec \theta = \dfrac{Hypotenuse}{Perpendicular}} \\\\ \sf{\star \sec \theta = \dfrac{Hypotenuse}{Base}}\\\\ \sf{\star \cot \theta = \dfrac{Base}{Perpendicular}} \end{array}}$

Then,

⠀⠀⠀⠀⠀$\sf{\tan \theta = \dfrac{7}{24}= \dfrac{Perpendicular} { Base} }$

Therefore,

• Perpendicular of Right Angled Triangle is 7 cm .

• Base of Right angled triangle is 24 cm .

⠀⠀⠀⠀⠀Finding Hypotenuse of Right angled triangle :

$\sf{\underline {\dag As, \:We \:Know\:that \::}}\\ \\ \bf{ By \:Pythagoras\:Theorem\::}$

$\underline {\boxed {\sf{\star (Perpendicular)^{2} + Base^{2} = ( Hypotenuse)^{2} }}}$

⠀⠀⠀⠀⠀⠀$\underline {\bf{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}$

⠀⠀⠀⠀⠀$:\implies \tt{ 7^{2} + 24^{2} = Hypotenuse ^{2}}$

⠀⠀⠀⠀⠀$:\implies \tt{ 49 + 576 = Hypotenuse^{2}}$

⠀⠀⠀⠀⠀$:\implies \tt{ Hypotenuse ^{2} = 625}$

⠀⠀⠀⠀⠀$:\implies \tt{ Hypotenuse = \sqrt{625}}$

⠀⠀⠀⠀⠀$\underline {\boxed{\pink{ \mathrm {  Hypotenuse = 25\: cm}}}}\:\bf{\bigstar}$

Therefore,

⠀⠀⠀⠀⠀$\underline {\therefore\:{ \mathrm {  Hypotenuse \:of\:Right \:Angled \:triangle \:is\:25\: cm}}}$

❒ Finding value of $\bf{\sin \theta }$ by using found values :

⠀⠀⠀⠀⠀$\sf{\underline {\dag As, \:We \:Know\:that \::}}$

⠀⠀⠀⠀⠀$\sf{\sin \theta = \dfrac{Perpendicular} { Hypotenuse} }$

Where ,

• Perpendicular of Right Angled Triangle is 7 cm .

• Hypotenuse of Right angled triangle is 25 cm .

Therefore,

⠀⠀⠀⠀⠀$\sf{\sin \theta = \dfrac{7}{25}= \dfrac{Perpendicular} { Hypotenuse} }$

⠀⠀⠀⠀⠀$\underline {\boxed{\pink{ \mathrm { Hence,\:The\:Value \:of\:\sin \theta = \dfrac{7}{25}\: }}}}\:\bf{\bigstar}$

❒ Finding value of $\bf{\sec \theta }$ by using found values :

⠀⠀⠀⠀⠀$\sf{\underline {\dag As, \:We \:Know\:that \::}}$

⠀⠀⠀⠀⠀$\sf{\sec \theta = \dfrac{Hypotenuse} { Base} }$

Where ,

• Base of Right Angled Triangle is 24 cm .

• Hypotenuse of Right angled triangle is 25 cm .

Therefore,

⠀⠀⠀⠀⠀$\sf{\sec \theta = \dfrac{25}{24}= \dfrac{Hypotenuse} { Base} }$

⠀⠀⠀⠀⠀$\underline {\boxed{\pink{ \mathrm { Hence,\:The\:Value \:of\:\sec \theta = \dfrac{25}{24}\: }}}}\:\bf{\bigstar}$

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