SSC CPO SI and ASI 2012 Paper

Question : 141 of 200

Marks: +1, -0

If sec (7θ + 28) = cosec (30 – 3θ) then the value of θ is

Solution:

(1)

\sec \ (7 \θ+28^{°}\ )=\cosec (30^{\°}-3 \θ)

=\sec \ \{90^{\°}-\ (30^{\°}-3 \θ)\ \}

7 \θ+28^{\°}=90^{\°}-30^{\°}+3 \θ

4 θ=60^{\°}-28

⇒θ=8^{\°}

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