65° 24' 30"

x =\({n\pi\over 3}+{\pi\over12},n \in z\)

2

\(cos \theta={4\over5},tan \theta={3\over4},sec\theta={5\over4},cot\theta {4\over3},cosec\theta={5\over 3}\)

\({\pi\over 20},{9\pi\over 20}\)

\({1237\over 21600}\pi\)

Here, x+ 2x+3x=180⁰ \(\Rightarrow \)x=30⁰
\(\therefore \) \(\angle \)A=30⁰, \(\angle \)B=60⁰, \(\angle \)C=90⁰
\(\therefore \)\(\angle \)A:\(\angle \)B:\(\angle \)C=1:\(\sqrt { 3 } \):2

\(\angle\)C=90⁰

Given cosec \(A=\frac { x }{ y } \)
We known that, cot²A = cosec² A-1
\({ cot }^{ 2 }A=\left( \frac { x }{ y } \right) ^{ 2 }-1=\frac { { x }^{ 2 } }{ { y }^{ 2 } } -1=\frac { { x }^{ 2 }-{ y }^{ 2 } }{ { y }^{ 2 } }\)
\(\Rightarrow \quad cot\quad A=\pm \frac { \sqrt { { x }^{ 2 }-{ y }^{ 2 } } }{ y } \)

\(sin^{ 2 }\frac { \pi }{ 6 } +cos^{ 2 }\frac { \pi }{ 6 } +sin^{ 2 }\frac { \pi }{ 4 } =\left( \frac { 1 }{ 2 } \right) ^{ 2 }+\left( \frac { \sqrt { 3 } }{ 2 } \right) ^{ 2 }+\left( \frac { 1 }{ \sqrt { 2 } } \right) ^{ 2 }\)
\(=\frac { 1 }{ 4 } +\frac { 3 }{ 4 } +\frac { 1 }{ 2 } =\frac { 1+3+2 }{ 4 } =\frac { 6 }{ 4 } =\frac { 3 }{ 2 } \)

We have
\( { -47 }^{ o }30'=-\left[ { 47 }^{ 0 }+\left( \frac { 30 }{ 60 } \right) ^{ o } \right] \quad \left[ \therefore 1'=\left( \frac { 1 }{ 60 } \right) ^{ o } \right] \)
\(=-\left[ { 47 }^{ o }+\left( \frac { 1 }{ 2 } \right) ^{ o } \right] =-\left( \frac { 95 }{ 2 } \right) ^{ o }\)
Now, required radian measure \(=\frac { \pi }{ 180 } \times \) Degree measure
\(=-\left( \frac { \pi }{ 180 } \times \frac { 95 }{ 2 } \right) rad=\frac { -19\pi }{ 72 } rad\)
Hence, radian measure of -47^o 30' is \(-\frac { 19\pi }{ 72 } rad\) .

\(cos\quad A\quad =\quad \sqrt { 1-\frac { 9 }{ 25 } } =\frac { 4 }{ 5 } ,\)
\(sin\quad B=\quad -\sqrt { 1-\frac { 144 }{ 169 } } =-\frac { 5 }{ 13 }\)
\(cos(A+B)\quad =\quad -\frac { 33 }{ 65 } \)

Here sin x =  =\({1\over4}\),x in quadrant II.
\(\therefore \ cos^2 x=1-sin^2 x\)
\(\Rightarrow\) cos²x= 1- \(({1\over4})^2=1-{1\over16}={15\over16}\)
\(\therefore cos x=\pm {\sqrt{15}\over4}\)
But x lies in second quadrant.
\(\therefore cos x=- {\sqrt{15}\over4}\)
Also \({x\over2}\)
So \({x\over2}\) lies in first quadrant.
\(\therefore \) sin\(x\over2\) ,cos\(x\over2\)and tan\(x\over2\)are all positive.
Now cos\({x\over2}={\sqrt{1+cos \ x\over 2}}=\sqrt{1+{\sqrt{15}\over4}\over2}\)
\(=\sqrt{4-\sqrt{15}\over 8}={\sqrt{8-2\sqrt{15}\over4}}\)
sin\({x\over2}={\sqrt{1-cos \ x\over 2}}=\sqrt{1+{\sqrt{15}\over4}\over2}\)
\(=\sqrt{4+\sqrt{15}\over 8}={\sqrt{8+2\sqrt{15}\over4}}\)
tan\(x\over2\) \(={sin {x\over2}\over cos {x\over2}}={{\sqrt{8+2\sqrt{15}}\over4}\over{\sqrt{8-2\sqrt{15}}\over4}}\)
\(={\sqrt{4+\sqrt{15}}\over\sqrt{4-\sqrt{15}}}\times {\sqrt{4+\sqrt{15}}\over\sqrt{4+\sqrt{15}}}\)
\(={{4+\sqrt{15}}\over\sqrt{16-16}}=4+\sqrt{15}\)

\(1\over8\)