MCQ
The distance between the points $(\text{a}\cos25^\circ,0)$ and $(0,\text{a}\cos65^\circ)$ is :
- ✓$a$
- B$2a$
- C$3a$
- DNone of these.
✓
Answer
Correct option: A.
$a$
Distance between $(\text{a}\cos25^\circ,0)$ and $(0,\text{a}\cos65^\circ)$
$=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
$=\sqrt{(0-\text{a}\cos25^\circ)^2+(\text{a}\cos65^\circ-0)^2}$
$=\sqrt{\text{a}^2\cos^225^\circ+\text{a}^2\cos^265^\circ}$
$=\sqrt{\text{a}^2[\cos^225^\circ+\cos^265^\circ]}$
$=\text{a}\sqrt{\cos^2(90^\circ-65^\circ)+\cos^265^\circ}$
$=\text{a}\sqrt{\sin^265^\circ+\cos^265^\circ}$
$=\text{a}(\sqrt{1})=\text{a}$
$=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
$=\sqrt{(0-\text{a}\cos25^\circ)^2+(\text{a}\cos65^\circ-0)^2}$
$=\sqrt{\text{a}^2\cos^225^\circ+\text{a}^2\cos^265^\circ}$
$=\sqrt{\text{a}^2[\cos^225^\circ+\cos^265^\circ]}$
$=\text{a}\sqrt{\cos^2(90^\circ-65^\circ)+\cos^265^\circ}$
$=\text{a}\sqrt{\sin^265^\circ+\cos^265^\circ}$
$=\text{a}(\sqrt{1})=\text{a}$
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