Class X Session 2023-24 Question 11

 MATH SAMPLE QUESTION PAPER

Class X Session 2023-24

MATHEMATICS STANDARD (Code No.041) 

Question 11

\begin{flalign} & Given \; that \; \sin\theta \; = \; \frac{a}{b},then \; \cos\theta\ is\;? &\\ \end{flalign}

\begin{flalign} (a)\;\;& \frac{b}{\sqrt{b^2 - a^2}}\\ (b)\;\;& \frac{b}{a} \\ (c)\;\;& \frac{\sqrt{b^2 - a^2}}{b}\\ (d)\;\;& \frac{a}{\sqrt{b^2 - a^2}}&\\ \end{flalign}

Explanation : 

\begin{flalign} & Given \; that \; \sin\theta \; = \; \frac{a}{b},to \; find \;the \; \cos\theta &\\ \end{flalign}

The Pythagorean identity states that in a right triangle, the square of the length of the hypotenuse (c) is equal to the sum of the squares of the lengths of the other two sides (𝑎 and 𝑏). Mathematically, it is represented as 

\[ \sin^2\Theta + \cos^2\Theta = 1, -----(a) \] 

\begin{flalign} & Since \; we \; know\; that\; \sin\theta \; = \; \frac{a}{b} &\\ \end{flalign}

we can substitute this value into the Pythagorean identity

\[ \Rightarrow  \frac{a^2}{b^2} + \cos^2\theta = 1 \] 

\[ OR \;  \cos^2\theta = 1 - \frac{a^2}{b^2} \]

Taking the square root of both sides

\[  \cos^2\theta =  \pm \sqrt{{1-}\frac{a^2}{b^2}} \]

Guess the Option and comment below  👇