Question

what is the answer of this question?

Answer 1

first find the value of sec and tan and sin

Answer 2

cosQ=27/29 cosQ= B /H
base= 27
hypotenuse= 29
perpendicular= ?
by Pythagoras theorem

(h)square = p square + b square
29 square= p square +27 square
841- 729=p square

$p = 8 \sqrt{3}$
secQ = H /B
tan Q= P /B
sin Q=P/ H
puts all the value
$29 \frac{27 \frac{8 \sqrt{3 } - \frac{8 \sqrt{3} } 29$
And solve