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Model Paper 6 — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSModel Paper 65 Marks
Question
If $\alpha, \beta$ are two different values of x lying between 0 and $2 \pi$ which satisfy the equation $6 \cos x +8 \sin x =9$, find the value of $\sin (\alpha+\beta)$
✓

Answer

We have to find the value of $\sin (\alpha+\beta)$
It is given that
6 cos x + 8 sin x = 9 
$\begin{array}{l}\Rightarrow 6 \cos x=9-8 \sin x \\ \Rightarrow 36 \cos ^2 x=(9-8 \sin x)^2 \\ \Rightarrow 36\left(1-\sin ^2 x\right)=81+64 \sin ^2 x-144 \sin x \\ \Rightarrow 100 \sin ^2 x-144 \sin x+45=0\end{array}$
Now, $\alpha$ and $\beta$ are the roots of the given equation; 
therefore, $\cos \alpha$ and $\cos \beta$ are the roots of the above equation. 
$\Rightarrow \sin \alpha \sin \beta=\frac{45}{100}$
(Product of roots of a quadratic equation $a x^2+b x+c=0$ is $\frac{c}{a}$ 
Again, 6 cosx + 8 sinx = 9 
$\begin{array}{l}\Rightarrow 8 \sin x=9-6 \cos x \\ \Rightarrow 64 \sin ^2 x=(9-6 \cos x)^2 \\ \Rightarrow 64\left(1-\cos ^2 x\right)=81+36 \cos ^2 x-108 \cos x \\ \Rightarrow 100 \cos ^2 x-108 \cos x+17=0\end{array}$
Now, $\alpha$ and $\beta$ are the roots of the given equation; 
therefore, $\sin \alpha$ and $\sin \beta$ are the roots of the above equation. 
Therefore, $\cos \alpha \cos \beta=\frac{17}{100}$
Hence, $\cos (\alpha+\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta$
$\begin{array}{l}=\frac{17}{100}-\frac{45}{100} \\ =-\frac{28}{100} \\ =-\frac{7}{25} \\ \sin (\alpha+\beta)=\sqrt{1-\cos ^2(\alpha+\beta)} \\ =\sqrt{1-\left(\frac{-7}{25}\right)^2} \\ =\sqrt{\frac{576}{625}} \\ =\frac{24}{25}\end{array}$

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