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Question
CBSEENMA10008287
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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

Solution

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°.


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.

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Question
CBSEENMA10008288
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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. 

Solution

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

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.

Question
CBSEENMA10008289
Wired Faculty App

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? 

Solution

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

Question
CBSEENMA10008290
Wired Faculty App

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.

Solution

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

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.

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