Trigonometric Equations Question 322
Question: If the line segment joining the points $ A(a,,b) $ and $ B(c,,d) $ subtends an angle $ \theta $ at the origin, then $ \cos \theta $ is equal to
[IIT 1961]
Options:
A) $ \frac{ab+cd}{\sqrt{(a^{2}+b^{2}),(c^{2}+d^{2})}} $
B) $ \frac{ac+bd}{\sqrt{(a^{2}+b^{2}),(c^{2}+d^{2})}} $
C) $ \frac{ac-bd}{\sqrt{(a^{2}+b^{2}),(c^{2}+d^{2})}} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- Here $ {{(AB)}^{2}}={{(a-c)}^{2}}+{{(b-d)}^{2}} $ $ {{(OA)}^{2}}={{(a-0)}^{2}}+{{(b-0)}^{2}}=a^{2}+b^{2} $ and $ {{(OB)}^{2}}=c^{2}+d^{2} $ Now from triangle $ AOB,\cos \theta =\frac{{{(OA)}^{2}}+{{(OB)}^{2}}-{{(AB)}^{2}}}{2OA.OB} $ $ =\frac{a^{2}+b^{2}+c^{2}+d^{2}-{{{(a-c)}^{2}}+{{(b-d)}^{2}}}}{2\sqrt{a^{2}+b^{2}}.\sqrt{c^{2}+d^{2}}} $ $ =\frac{ac+bd}{\sqrt{(a^{2}+b^{2})(c^{2}+d^{2})}} $ .
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