Question

In AB is a 6 m high pole and CD is a ladder inclined at an angle 60 server to the horizontal and reaches up to a point D of pole .If AD=2.54m , find the length of the ladder(use rout 3=1.73)

Answer 1

AB IS THE POLE OF 6M
CD IS THE LADDER MAKING AN ANGLE OF 60

AB =6
AD=2.54

So,DB= 6 - 2.54
=3.46m

Now in ∆DBC,
SIN 60=DB/DC
✓3/2 =3.46/DC
✓3DC =3.46 ×2
1.73DC =6.92
DC=6.92/1.73
DC=4m
Length of ladder is 4m

Answer 2

Answer:

The length of the ladder is 4 m.

Explanation:

• AB is a 6 m high pole which is vertically upward.
• CD is the length of the ladder which touches AB at D making 60⁰ angle with ground.
• The length of AD is 2.54 m .
• From the figure we can say that

AB = AD + DB

6 = 2.54 + DB

or, DB = ( 6-2.54 ) m

DB = 3.46 m

• ∆ DBC is a right angled triangle with right angle at B.
• BC is the base of the triangle, DB is perpendicular of triangle and CD is hypotenuse of the triangle.
• To find the length of CD, we have to establish a relationship between CD and DB.
• We know that

sin ∅ = perpendicular/ hypotenuse

• Using this relation in ∆ DBC, we get

$\sin( {60}^{0} ) = \frac{db}{cd} \\ \frac{ \sqrt{3} }{2} = \frac{3.46}{cd} \\ \frac{1.73}{2} = \frac{3.46}{cd} \\ cd = \frac{3.46 \times 2}{1.73} \\ cd = 2 \times 2 \\ cd = 4 \: m$

• Hence, the length of the ladder CD is 4 m.