Question

In figure AB is a 6 m
high pole and CD is
a ladder inclined at
an angle of 60° to
the horizontal and
reaches up to a
point D of pole. If
AD = 2.54 m. find
the length of the
ladder.
(Use√3 = 1.72)
​

Answer 1

Answer:

11.98

bd=3.46

tan 60=√3/1

bd=√3

cb=1

so dc =2

(3.36/√3)*2=ans

Answer 2

Answer:

4 m

Step-by-step explanation:

Sin ∅ = $\frac{Opposite}{Hypotenuse}$                                                                       2.54 - 6 = - 3.46

Sin 60 ° = $\frac{DB}{DC}$

$\frac{\sqrt{3} }{2} = \frac{-3.64}{DC} \\\\DC = \frac{2(3.64)}{1.73}$

DC = 4 m

If this answer is helpful mark me as Brainliest