Questions
If the segment joining the points
a
cosθ=ac+bd(a2+b2)(c2+d2)
b
sinθ=ac+bd(a2+b2)(c2+d2)
c
cosθ=ac−bd(a2+b2)(c2+d2)
d
sinθ=ac−bd(a2+b2)(c2+d2)
detailed solution
Correct option is A
Let O be the origin. Then OA2=a2+b2,OB2=c2+d2 and AB2=(c−a)2+(d−b)2 Using cosine formula in ΔOAB, we haveAB2=OA2+OB2−2.OA.OBcosθ⇒ (c−a2)+(d−b)2=a2+b2+c2+d2−2a2+b2c2+d2cosθ⇒ c2+a2−2ac+d2+b2−2bd=a2+b2+c2+d2−2a2+b2c2+d2cosθ⇒ 2(ac+bd)=2a2+b2c2+d2cosθ⇒ cosθ=ac+bda2+b2c2+d2.Talk to our academic expert!
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