Question

if sec a=5/4 then evaluate sinalpha/secalpha

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{\frac{sin \: \alpha }{sec \: \alpha } =\frac{12}{25}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies sec \: \alpha = \frac{5}{4 } \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies \frac{sin \: \alpha }{sec \: \alpha } = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies sec \: \alpha = \frac{5}{4} \\ \\ \tt: \implies \frac{hypotenuse}{base} = \frac{5}{4} \\ \\ \bold{From \: phythagoras \: theorem} \\ \tt: \implies {h}^{2} = {p}^{2} + {b}^{2} \\ \\ \tt: \implies {5}^{2} = {p}^{2} + {4}^{2} \\ \\ \tt: \implies 25 - 16 = {p}^{2} \\ \\ \tt: \implies p = \sqrt{9} \\ \\ \tt: \implies p = 3 \\ \\ \bold{For \: finding \: value} \\ \tt: \implies \frac{sin \: \alpha }{sec \: \alpha } = \frac{ \frac{p}{h} }{ \frac{h}{b}} \\ \\ \tt: \implies \frac{sin \: \alpha }{sec \: \alpha } = \frac{ \frac{3}{5} }{ \frac{5}{4} } \\ \\ \tt: \implies \frac{sin \: \alpha }{sec \: \alpha } = \frac{3}{5} \times \frac{4}{5} \\ \\ \green{\tt: \implies \frac{sin \: \alpha }{sec \: \alpha } = \frac{12}{25} }$

Answer 2

$\mathtt{ \huge{ \fbox{Solution :)}}}$

Given ,

Sec(Φ) = 5/4 = H/B

By pythagoras theorem ,

(H)² = (B)² + (P)²

(5)² = (4)² + (P)²

25 = 16 + (P)²

(P)² = 9

P = √9

P = 3 units

[ ignore the negative value of P because length can't be negative ]

Now ,

$\sf \hookrightarrow \frac{Sin(Φ) }{Sec(Φ) } = \frac{ \frac{P}{H } }{ \frac{H}{B} } \\ \\ \sf \hookrightarrow \frac{Sin(Φ) }{Sec(Φ) } = \frac{PB}{ {(H)}^{2} } \\ \\ \sf \hookrightarrow \frac{Sin(Φ) }{Sec(Φ) } = \frac{3 \times 4}{ {(5)}^{2} } \\ \\ \sf \hookrightarrow \frac{Sin(Φ) }{Sec(Φ) } = \frac{12}{25}$

Hence , Sin(Φ)/Sec(Φ) = 12/15

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