[tex]\large\huge\green{\sf{Answer:-}}[/tex]
- option c is correct answer
[tex]\huge{\underline{\underline{\boxed{\sf{\green{Explanation:-}}}}}}[/tex]
[tex]\large\huge\green{\sf{Given:-}}[/tex]
- ABC a right triangle:-
- hypotenuse= AC= x
- base= 12 unit
- perpendicular=AB = y
[tex]\large\huge\green{\sf{ToFind:-}}[/tex]
- Hypotenuse=AC = x=??
- perpendicular=AB = y=??
[tex]\large\huge\green{\sf{FormulaRequired:-}}[/tex]
- [tex] sec Q= \frac{hypotense}{base} [/tex]
- value of sec 30°= 2/√3
- [tex] sin Q= \frac{perpendicular}{hypotenuse} [/tex]
- value of sin30° = 1/2
[tex]\large\huge\green{\sf{Solution:-}}[/tex]
[tex] \red {\mathbb{ \underline { \tt By \: Using \: Trigonometric\: Formula:-}}}[/tex]
[tex]for \: x : - \\ \sec(Q) = \frac{h}{b} \\ sec(30 {}^{o} ) = \frac{x}{12} \\ \sec(30) = \frac{2}{ \sqrt{3} } \\ \frac{2}{ \sqrt{3} } = \frac{x}{12} \\ x = \frac{12 \times 2}{ \sqrt{3} } \\ x = \frac{24}{ \sqrt{3} } \: \\ by \: rationalise \: denomator \\ x = \frac{24 \times \sqrt{3} }{ \sqrt{3 } \times \sqrt{3} } \\ x = \frac{24 \sqrt{3} }{3} \\ x = 8 \sqrt{3} [/tex]
[tex]for \: y : - \\ sinQ = \frac{p}{h} \\ sin30 {}^{o} = \frac{y}{8 \sqrt{3} } \\ sin30 {}^{o} = \frac{1}{2} \\ \frac{1}{2} = \frac{y}{8 \sqrt{3} } \\ y = \frac{8 \sqrt{3} }{2} \\ y = 4 \sqrt{3} [/tex]
