MCQ
The distance between the points $\text{A}(\text{p}\sin 25^\circ,0)$ and $\text{B}(\text{0, p }\sin 65^\circ)$ is :
- A$0$ units.
- ✓$p$ units.
- C$1$ units.
- D$p^2$ units.
✓
Answer
Correct option: B.
$p$ units.
The distance between point $\text{A(p }\sin25^\circ,0)$ and point $\text{B(p }\sin65^\circ)$
$\text{AB}\sqrt{(0-\text{p}\sin25^\circ)^2+(0-\text{p}\sin25^\circ)}$
$=\sqrt{\text{p}^2\sin^225^\circ+\text{p}^2\sin^265^\circ}$
$=\text{p}\sqrt{\sin^225^\circ+\sin^2(90^\circ-25^\circ)}$
$=\text{p}\sqrt{\sin^225^\circ+\cos^\circ25^\circ}\ [\because\sin(90^\circ-\theta)=\cos\theta]$
$=\text{p units}$
$[\because\cos^2\theta+\sin^2\theta=1]$
$\text{AB}\sqrt{(0-\text{p}\sin25^\circ)^2+(0-\text{p}\sin25^\circ)}$
$=\sqrt{\text{p}^2\sin^225^\circ+\text{p}^2\sin^265^\circ}$
$=\text{p}\sqrt{\sin^225^\circ+\sin^2(90^\circ-25^\circ)}$
$=\text{p}\sqrt{\sin^225^\circ+\cos^\circ25^\circ}\ [\because\sin(90^\circ-\theta)=\cos\theta]$
$=\text{p units}$
$[\because\cos^2\theta+\sin^2\theta=1]$
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