Question

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

• A. < 8
• B. < 7
• C.
• D. < 7

Answer 1

The correct option is B < 7

$\theta$ is an acute angle so $0^{\circ }\le \theta <{90}^{\circ }$
$\therefore 0\le \frac{p-6}{8-p}$ < 1 $\Rightarrow 0\le (p-6)$ < $(8-p)\Rightarrow 6\le p$ < 7.