1.

The plane of dip circle is set in thegeographical meridian and the apparent dip is theata it is then set in a verticalplane prependicular to thegeopgraphical meridian the apparent dip becomes theta_(2)the angle of declinationalpha is given by

Answer»

`tan alpha=sqrt(tan theta_(1)xxtantheta_(2))`
`tan alpha=sqrt(tan theta_(1)xxtan theta_(2))^(2)`
`tanalpha=tan theta_(1)//tan theta_(2)`
`tan alpha=tan thetasqrt(tan)theta_(1)`

Solution :Here in geographcal MERIDIAN
`tan theta_(1)=(V)/(H COS alpha)`
`tan theta_(2)=(V)/(H cos (90^(@)-alpha)) =(v)/(Hsin alpha)`
`THEREFORE (tantheta_(1))/(tan theta_(2))=(sin alpha)/(cos alpha)=tan alpha`


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