Step-by-step explanation:
given,
tan thita =8/9
or,tan thita=p/b
now, h=
[tex] \sqrt{p {}^{2} + b {}^{2} } [/tex]
[tex] = \sqrt{8 {}^{2} + 9 {}^{2} } [/tex]
[tex] = \sqrt{145} [/tex]
therefore, h=
[tex] \sqrt{145} [/tex]
now, cos thita =b/h
so, cos thita =
[tex]9 \div \sqrt{145} [/tex]
[tex] = 9 \sqrt{145} \div \sqrt{145} [/tex]
......ans........
hope its helpful
