Question

Q:-If tan θ =12/5 ,then show that 5 cos θ + 12 sin θ = 13​

Answer 1

$\huge{\bold☘}\mathfrak\pink{\bold{\underline{{ ℘ɧεŋσɱεŋศɭ}}}}{\bold☘}$

$\red{\bold{\underline{\underline{QUESTION:-}}}}$

Q:-If tan θ =12/5 ,then show that 5 cos θ + 12 sin θ = 13

$\huge\tt\underline\blue{Answer }$

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$⟹\bold{tan \theta = \frac{p}{b} = \frac{12}{5}}$

$⟹ \bold{{h}^{2} = {p}^{2} + {b}^{2}}$

$⟹\bold{ {h}^{2} = 144 + 125 = 169}$

$⟹\bold{h = \sqrt{169} = 13}$

$⟹\bold{cos \theta = \frac{b}{h} = \frac{5}{13}}$

$⟹\bold{sin \theta = \frac{p}{h} = \frac{12}{13} }$

$\bold{Now \: 5 \times \frac{5}{13} + 12 \times \frac{12}{13} = 13}$

$⟹\bold{ \frac{25}{13} + \frac{144}{13} = 13}$

$⟹\bold{ \frac{169}{13} = 13}$

$\bold{13 = 13}$

$\bold{\red{From\: above\: it\: is\: proved \:that [tex]tan\theta=12/5}}$

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Answer 2

Let us consider , 5cosΘ + 12sinΘ = 13

Dividing all sides by 13...

We get,

$\bf\red{\underline{}} \frac{5cos \alpha + 12sin \alpha }{13} = 1 \\ \\ \bf\red{\underline{}} \frac{5cos \alpha }{13} + \frac{12sin \alpha }{13} = 1 \\ \\ \bf\large\red{\underline{ {sin}^{2} \alpha + {cos}^{2} \alpha = 1}}$

Therefore,

sinΘ = 12/13

cosΘ = 5/13

➜tanΘ = sinΘ/cosΘ

➜

$= > tan \: \alpha = \bf\large\red{\underline{}} \frac{ \frac{12}{13} }{ \frac{5}{13} } \\ \\ = \bf\large\red{\underline{}}> tan \alpha = \frac{12}{5}$

tanΘ = 12/5