find the value of angle whose sine is given as `sintheta = (sqrt 3/2)`

Find the angles of the triangle whose sides are 3+ sqrt3, 2sqrt3 and sqrt6.

Problem on sine rule Type:-1(i) In a DeltaABC ; If a=2;b=3 and sinA=2/3 ;find /_B (ii) In a DeltaABC ; the angle of a triangle are in AP ; It is being given that b:c=sqrt3:sqrt2

Using the given figure , find the value of angle A , if : (i) x= y (ii) x= sqrt3 y (iii) sqrt3, x=y

Write the acute angle theta satisfying sqrt(3)sintheta=costheta .

Find the angle between the lines whose direction cosines are (-(sqrt3)/(4), (1)/(4), -(sqrt3)/(2)) and (-(sqrt3)/(4), (1)/(4), (sqrt3)/(2)) .

find general value of theta : sintheta=-1/2 and costheta=-sqrt(3)/2

If 3tantheta=4 , find the value of (4costheta-sintheta)/(2costheta+sintheta)

Find the general solutions of the following equation: sintheta=(sqrt(3))/2

If 3cottheta=4, find the value of (4costheta-sintheta)/(2costheta+sintheta) .

Find the geneal values of theta from the following equations: i) sintheta=sqrt(3)/2 , ii) costheta=1/sqrt(2) , iii) tantheta=sqrt(3) , iv) sectheta=2/sqrt(3) , v) cottheta=1/sqrt(3) , vi) "cosec"theta=sqrt(2)