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This topic is not avialable

Raghavendra

· started a discussion

· 1 Months ago

Wrong answer (Question:99510.0)

As cos© = b-c/a

This topic is not avialable

Ashish Sajwan

· started a discussion

· 1 Months ago

Wrong answer (Question:99510.0)

Option C satisfies

Prashant Kushwaha

· started a discussion

· 1 Months ago

Wrong answer (Question:99510.0)

Hey the answer is different and it's also not present in option and this answer and solution is totally wrong. Please check it out

Question:
If a = b sin θ + c and b = a cos θ + c, then
Options:
A) \(\cfrac{(a-b)^2}{c^2}+\cfrac{(b-a)^2}{c^2}=1\)
B) \(\left ( \cfrac {a-c}{a} \right )^2+\left ( \cfrac {b-c}{a} \right )^2=1\)
C) \(\left ( \cfrac {a-c}{c} \right )^2+\left ( \cfrac {b-c}{c} \right )^2=1\)
D) \(\cfrac{(a-c)^2}{b^2}+\cfrac{(b-c)^2}{a^2}=1\)
Solution:
Ans: (d) a = b sin θ + c ⇒ \(\cfrac{a-c}{b}=\) sinθ   …(1)

b = a cos θ + c ⇒ \(\cfrac{b-c}{a}=\) cosθ    …(2)

Squaring and adding (1) and (2)

sin2 θ +cos2 θ = \(\left ( \frac {a-c}{b} \right )^2+ \left ( \frac {b-c}{a} \right )^2 =1\)

 \(\cfrac{(a-c)^2}{b^2}+\cfrac{(b-c)^2}{a^2}=1\)

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