Question

sinA=3/4,calculate secA

Answer 1

$\huge{\mathfrak{\green{\star{Answer}}}}$

SinA = 3/4 = P/H

sec = H/B

$h {}^{2} = b {}^{2} + p {}^{2} \\ (4) {}^{2} = b{}^{2} + (3) {}^{2} \\ 16 = b {}^{2} + 9 \\ 16 - 9 = b {}^{2} \\ 7 = b {}^{2} \\ \sqrt{7} = b$

Sec A = H/B = 4/√7

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Answer 2

Answer:

Sec A = 4/√7

Step-by-step explanation:

Find all the 3 sides of right angled triangle by Pythagorous theorem.

Sin A = $\frac{3}{4}$ = $\frac{perpendicular}{hypotenuse}$

AC² = AB² + BC²

4² = AB² + 3²

AB² = 16 - 9

AB = √7

Base/adj = √7

so,

Sec A = $\frac{hypotenuse}{Base}$

Sec A = 4/√7

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