Question

If $A=\sin {46}^{\circ }\sin {20}^{\circ }$; $B=\cos {45}^{\circ }\cos {12}^{\circ }$; $C=\cos {66}^{\circ }+\sin {84}^{\circ }$, then descending order of these values is

• A. $C,A,B$
• B. $C,B,A$
• C. $A,C,B$
• D. $A,B,C$

Answer 1

Answer: (B)

$C,B,A$

$A=\sin 45\cdot \sin 12=\frac{1}{\sqrt{2}}\sin 12$

$B=\cos 45\cdot \cos 12=\frac{1}{\sqrt{2}}\cos 12$

$C=\cos 66+\sin 84=\cos 66+\sin (90-6)$

$=\cos 66+\cos 6=2\cos 36\cos 30=\sqrt{3}\cos 36$

For $\theta ϵ(0,45)\Rightarrow \cos \theta >\sin \theta$

$\Rightarrow B>A$

Now

$\cos 12>\cos 36\Rightarrow \frac{1}{\sqrt{2}}\cos 12<\cos 36$

$\Rightarrow \frac{1}{\sqrt{2}}\cos 12<\sqrt{3}\cos 36\Rightarrow C>B$

Therefore $C>B>A$