If tan thita =7/24 then find the value of sin thita and sec thita
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If tan thita =7/24 then find the value of sin thita and sec thita
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Answer:
[tex]\displaystyle{{\boxed{\red{\pink{\sf\:\sin\:\theta\:=\:\dfrac{7}{25}}}}}}[/tex]
[tex]\displaystyle{\boxed{\blue{\sf\:\sec\:\theta\:=\:\dfrac{25}{24}}}}[/tex]
Step-by-step-explanation:
We have given that,
[tex]\displaystyle{\sf\:\tan\:\theta\:=\:\dfrac{7}{24}}[/tex]
We have to find the value of [tex]\displaystyle{\sf\:\sin\:\theta\:\&\:\sec\:\theta}[/tex]
Now, we know that,
[tex]\displaystyle{\pink{\sf\:\tan\:\theta\:=\:\dfrac{Opposite\:side}{Adjacent\:side}}}[/tex]
[tex]\displaystyle{\implies\sf\:\dfrac{7}{24}\:=\:\dfrac{Opposite\ side}{Adjacent\ side}}[/tex]
∴ Opposite side = 7 units
Adjacent side = 24 units
Now, we know that,
[tex]\displaystyle{\sf\:(\:Hypotenuse\:)^2\:=\:(\:Opposite\ side\:)^2\:+\:(\:Adjacent\:side\:)^2}[/tex]
[tex]\displaystyle{\implies\sf\:H^2\:=\:(\:7\:)^2\:+\:(\:24\:)^2}[/tex]
[tex]\displaystyle{\implies\sf\:H^2\:=\:49\:+\:576}[/tex]
[tex]\displaystyle{\implies\sf\:H^2\:=\:625}[/tex]
[tex]\displaystyle{\implies\sf\:H\:=\:\sqrt{625}}[/tex]
[tex]\displaystyle{\implies\boxed{\red{\sf\:Hypotenuse\:=\:25\:units}}}[/tex]
Now, we know that,
[tex]\displaystyle{\pink{\sf\:\sin\:\theta\:=\:\dfrac{Opposite\ side}{Hypotenuse}}}[/tex]
[tex]\displaystyle{\implies\sf\:\sin\:\theta\:=\:\dfrac{7}{25}}[/tex]
[tex]\displaystyle{\therefore\:\underline{\boxed{\red{\pink{\sf\:\sin\:\theta\:=\:\dfrac{7}{25}}}}}}[/tex]
Now, we know that,
[tex]\displaystyle{\blue{\sf\:\sec\:\theta\:=\:\dfrac{Hypotenuse}{Adjacent\ side}}}[/tex]
[tex]\displaystyle{\implies\sf\:\sec\:\theta\:=\:\dfrac{25}{24}}[/tex]
[tex]\displaystyle{\therefore\:\underline{\boxed{\blue{\sf\:\sec\:\theta\:=\:\dfrac{25}{24}}}}}[/tex]
Given : [tex]\sf{\tan \theta = \dfrac{7}{24}}\\[/tex]
Need To Find : The value of [tex]\sin \theta \:and\:\sec\theta [/tex]
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❍ Basic Formulas of Trigonometry is given by :
[tex]\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}}\\[/tex]
Then,
⠀⠀⠀⠀⠀[tex]\sf{\tan \theta = \dfrac{7}{24}= \dfrac{Perpendicular} { Base} }\\[/tex]
Therefore,
⠀⠀⠀⠀⠀Finding Hypotenuse of Right angled triangle :
[tex]\sf{\underline {\dag As, \:We \:Know\:that \::}}\\ \\ \bf{ By \:Pythagoras\:Theorem\::}\\[/tex]
[tex]\underline {\boxed {\sf{\star (Perpendicular)^{2} + Base^{2} = ( Hypotenuse)^{2} }}}\\[/tex]
⠀⠀⠀⠀⠀⠀[tex]\underline {\bf{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}\\[/tex]
⠀⠀⠀⠀⠀[tex]:\implies \tt{ 7^{2} + 24^{2} = Hypotenuse ^{2}}\\[/tex]
⠀⠀⠀⠀⠀[tex]:\implies \tt{ 49 + 576 = Hypotenuse^{2}}\\[/tex]
⠀⠀⠀⠀⠀[tex]:\implies \tt{ Hypotenuse ^{2} = 625}\\[/tex]
⠀⠀⠀⠀⠀[tex]:\implies \tt{ Hypotenuse = \sqrt{625}}\\[/tex]
⠀⠀⠀⠀⠀[tex]\underline {\boxed{\pink{ \mathrm { Hypotenuse = 25\: cm}}}}\:\bf{\bigstar}\\[/tex]
Therefore,
⠀⠀⠀⠀⠀[tex]\underline {\therefore\:{ \mathrm { Hypotenuse \:of\:Right \:Angled \:triangle \:is\:25\: cm}}}\\[/tex]
❒ Finding value of [tex]\bf{\sin \theta }[/tex] by using found values :
⠀⠀⠀⠀⠀[tex]\sf{\underline {\dag As, \:We \:Know\:that \::}}\\ \\ [/tex]
⠀⠀⠀⠀⠀[tex]\sf{\sin \theta = \dfrac{Perpendicular} { Hypotenuse} }\\[/tex]
Where ,
Therefore,
⠀⠀⠀⠀⠀[tex]\sf{\sin \theta = \dfrac{7}{25}= \dfrac{Perpendicular} { Hypotenuse} }\\[/tex]
⠀⠀⠀⠀⠀[tex]\underline {\boxed{\pink{ \mathrm { Hence,\:The\:Value \:of\:\sin \theta = \dfrac{7}{25}\: }}}}\:\bf{\bigstar}\\[/tex]
❒ Finding value of [tex]\bf{\sec \theta }[/tex] by using found values :
⠀⠀⠀⠀⠀[tex]\sf{\underline {\dag As, \:We \:Know\:that \::}}\\ \\ [/tex]
⠀⠀⠀⠀⠀[tex]\sf{\sec \theta = \dfrac{Hypotenuse} { Base} }\\[/tex]
Where ,
Therefore,
⠀⠀⠀⠀⠀[tex]\sf{\sec \theta = \dfrac{25}{24}= \dfrac{Hypotenuse} { Base} }\\[/tex]
⠀⠀⠀⠀⠀[tex]\underline {\boxed{\pink{ \mathrm { Hence,\:The\:Value \:of\:\sec \theta = \dfrac{25}{24}\: }}}}\:\bf{\bigstar}\\[/tex]
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