Question

Plz find 33,34,35,36
Plz fast

Answer 1

0u33.
in a triangle ABC in which /_B=90°
it is given that , 5tan=4
so,tan=4/5_______○1
but in triangle ABC
AB/BC=tan _______○2
from ○1 and○2
AB/BC=4/5
thus AB=4 and BC=5
by PGT - (AC)^2=(4)^2+(5)^2
(AC)^2=16+25
AC=underroot 41
so,sin=4/underroot 41
and cos=5/underroot 41
now u can get the answer by putting these values of sin and cos in the question.

34.
it is also similar to above question
here,
BC/AB=4/3
so,by pgt -
AC^2=4^2+3^2
Ac^2=16+9
AC=5
so, cosec =5/3
put these values in question and solve it.

35.sin (A-B)=1/2
sin (A-B)=sin30°
[it is because sin30=1/2]
so, (A-B)=30
A=30+B________○3
again, cos (A+B)=1/2
cos(A+B)=cos60°
[it is because cos60=1/2]
A+B=60
putting value of eq3
30+B+B=60
2B=30
B=15°
so,A=45°

36.
since A,B,C are angles of triangle ABC
therefore, A+B+C=180°
A+B=180-C_______○4
now, cosec [A+B/2]
=> cosec [180-C/2] (from4)
=> cosec [90-C/2]
=>sec C/2
[this is because
cosec (90- theta)=sec theta]

hope u got the solutions...!!!