Question

In Fig. 1, 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=1.73)

Answer 1

Hey user !

Here's the answer :)

Given :-

AB = 6 m
AD = 2.54 m

Hence,

AB - AD = BD

6 - 2.54 = BD

3.46 m = BD

Now,

In ∆ DBC,

$\sf{ \sin \: 60 \degree = \frac{BD}{DC} }$

$\sf{ \frac{ \sqrt{3} }{2} = \frac{3.46}{BD}}$

As per given,

√3 = 1.73

$\sf{1.73 \times BD = 3.46 \times 2}$

$\sf {BD = \frac{3.46 \times 2}{1.73}}$

$\sf{BD = 2 \times 2}$

$\sf{BD = 4}$

Therefore,
Length of the ladder is 4 m.

Thanks!

Answer 2

Answer:

4m

Step-by-step explanation:

Find the length of BD:

BD = 6 - 2.54 = 3.46 m

Find the length of the ladder:

Sin θ = opp/hyp

sin (60) = 3.46/CD

CD = 3.46 ÷ sin (60)

CD = 4 m

Answer: The length of the ladder is 4 m