(B) Since $\theta$ is an acute angle,we have $0 < \theta < 90^{\circ}$.Therefore,$0 < \sin\theta < 1$.Substituting the given expression: $0 < \frac{p

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(B) Since $\theta$ is an acute angle,we have $0 Therefore,$0 Substituting the given expression: $0 Case $1$: $\frac{p - 6}{8 - p} \ge 0$. Since the denominator $8 - p$ must be positive for $\sin\theta$ to be defined and positive,$p Case $2$: $\frac{p - 6}{8 - p} 0$,we have $p - 6 Combining these,we get $6 \le p < 7$.

Class 11 Mathematics Linear Inequalities Word problem of Linear inequalities

Difficult

If $\theta$ is an acute angle and $\sin\theta = \frac{p - 6}{8 - p}$,then $p$ must satisfy

A
$6 \le p < 8$
B
$6 \le p < 7$
C
$3 \le p \le 4$
D
$4 \le p < 7$

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Word problem of Linear inequalities

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If $\theta$ is an acute angle and $\sin\theta = \frac{p - 6}{8 - p}$,then $p$ must satisfy

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