zigya tab

The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.


Let A be the top and B be the foot of the tower AB, of height h metre. C be the point which is 30 m away from the tower i.e. BC = 30 m.
Now,
AB = h m
BC = 30 m
and    ∠ACB = 30°
In right triangle ABC, we have

tan space 30 degree space equals space AB over BC

Let A be the top and B be the foot of the tower AB, of height h metre
rightwards double arrow space space space fraction numerator 1 over denominator square root of 3 end fraction equals straight h over 30
rightwards double arrow space space space square root of 3 straight h end root space equals space 30
rightwards double arrow space space space space straight h space equals space fraction numerator 30 over denominator square root of 3 end fraction
rightwards double arrow space space space space straight h space equals space fraction numerator 30 over denominator square root of 3 end fraction straight X fraction numerator square root of 3 over denominator square root of 3 end fraction
space space space space space space space space space space space equals space fraction numerator 30 square root of 3 over denominator 3 end fraction space equals space 10 square root of 3
Hence, the height of the tower is 10 square root of 3 space straight m.

1076 Views

A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case? 


Case I :

Calculation of length of the slide below 5 years. Fig. (a)
Let AC be the slide of length l m and height of the slide AB of height 1.5 m. It is given that slide is inclimed at an angle of 30°.
i.e.,    ∠ACB = 30°
In right triangle ABC, we have

sin space 30 to the power of degree space equals space AB over AC
rightwards double arrow space space 1 half space equals space fraction numerator 1.5 over denominator straight l end fraction
rightwards double arrow space space space space straight l space space equals space space 3 space straight m space

Hence, the length of the slide for below 5 years is 3 m

Case II :

Calculation of the length of slide for elder children. Fig. (b)
Let DF be the slide of length m metres and DE be the height of the top of slide of height 3 m. It is given that slide is inclined at an angle of 60°.
i.e.,    ∠DFE = 60°
In right triangle DEF, we have

sin space 60 degree space equals space DE over DF
rightwards double arrow space space space fraction numerator square root of 3 over denominator 2 end fraction equals 3 over straight m
rightwards double arrow space space space space square root of 3 space straight m space space equals space 6
rightwards double arrow space space space space straight m space space equals space fraction numerator 6 over denominator square root of 3 end fraction
space space space space space space space space space space equals space fraction numerator 6 cross times space square root of 3 over denominator square root of 3 cross times square root of 3 end fraction space equals space 2 square root of 3 space straight m
Hence, the length of the slide for elder children is 2 square root of 3 space straight m

884 Views

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree. 


Let BD be the tree broken at point C such that the broken part CD takes the position CA and strikes the ground at A. It is given that AB = 8 m and ∠BAC = 30°.
Let BC = x metres and CD = CA = y metres
In right triangle ABC, we have

tan space 30 to the power of degree space equals space BC over AB
rightwards double arrow space space fraction numerator 1 over denominator square root of 3 end fraction equals straight x over 8
rightwards double arrow space space space straight x space equals space fraction numerator 8 over denominator square root of 3 end fraction equals fraction numerator 8 square root of 3 over denominator 3 end fraction
and space cos space 30 to the power of degree space equals space AB over AC
rightwards double arrow space space fraction numerator square root of 3 over denominator 2 end fraction space equals space 8 over AC
rightwards double arrow space space space AC space space equals space fraction numerator 16 over denominator square root of 3 end fraction

Let BD be the tree broken at point C such that the broken part CD tak

rightwards double arrow space space space space straight y space space space equals space fraction numerator 16 over denominator square root of 3 end fraction equals fraction numerator 16 square root of 3 over denominator 3 end fraction
Now, height of the tree

equals space left parenthesis straight x space plus space straight y right parenthesis space space metres
equals open square brackets fraction numerator 8 square root of 3 over denominator 3 end fraction plus fraction numerator 16 square root of 3 over denominator 3 end fraction close square brackets space metres
equals space fraction numerator 24 square root of 3 over denominator 3 end fraction space metres
equals 8 square root of 3 space metres
Hence, the height of the tree equals space 8 square root of 3 space straight m.

Let BD be the tree broken at point C such that the broken part CD tak

1331 Views

Advertisement

A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30° (see Fig. 9.11)

Fig. 9.11


Let AC be the rope whose length is 20 m, and AB be the vertical pole of height h m and the angle of elevation of A at point C on the ground is 30°.




Let AC be the rope whose length is 20 m, and AB be the vertical pole

Fig. 9.11
i.e.,    ∠ACB = 30°
In right triangle ABC, we have

sin space 30 to the power of degree space equals space AB over AC
rightwards double arrow space space space space 1 half space equals space straight h over 20
rightwards double arrow space space space space 2 straight h space equals space 20
rightwards double arrow space space space space space straight h space equals space 10 space straight m

Hence, the height of the pole is 10 m.

2146 Views

Advertisement

A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.


Let BC be the horizontal ground, and let A be the position of the kite which is at a height of 60 m. i.e., AB = 60 m.
It is given that
∠ACB = 60°
Now, in right triangle ABC, we have

sin space space 60 degree space space space equals space AB over AC
rightwards double arrow space space space fraction numerator square root of 3 over denominator 2 end fraction space equals space 60 over AC
rightwards double arrow space space space space AC space space space equals space fraction numerator 60 space straight X space 2 over denominator square root of 3 end fraction
rightwards double arrow space space space space AC space space space equals space fraction numerator 120 over denominator square root of 3 end fraction space straight X space fraction numerator square root of 3 over denominator square root of 3 end fraction
rightwards double arrow space space space space AC space space space equals space space 40 square root of 3 space space straight m
Hence, the length of string is 40 square root of 3 space straight m.

1406 Views

Advertisement