Math, asked by stevewith8jobs, 6 months ago

a 8m 40cm high vertical pole cast a shadow 4m 80cm.find at the same time.
a)the height of a pole which casts a shodow 4m long.
b)the length of the shadow cast by another pole 12m 25 cm high​

Answers

Answered by shagun6568
6

Step-by-step explanation:

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Answered by XxxRAJxxX
3

Given :-

  • Height of the pole (P) = 8m 40 cm
  • Length of the shadow (B) = 4m 80 cm

Solution:-

In Right angled Δ ABC formed,

Height of the pole, Perpendicular, P = 8.40 m

Length of the shadow, Base, B = 4.80 m

Since,

 \because \rm \tan \theta = \frac{P}{B}

Therefore,

 \therefore \mathrm{\tan \theta = \frac{8.40}{4.80}}

 \implies \mathrm{\tan \theta = \cancel{\frac{8.40}{4.80}}}

\implies \boxed{\bold{\mathrm{\tan \theta = \frac{3}{2}}}}

(a) To find :- The height of a pole which casts a shadow 4m long.

Solution:-

Length of the shadow, Base, B = 4m

Since,

 \because \rm \tan \theta = \frac{P}{B}

Therefore,

 \therefore \rm \tan \theta = \frac{P}{4}

Hence, the sun is casting the same angle of elevation of light, so angle of suspension of shadow formed of all poles is equals, i.e., \it \tan \theta is equal in both poles.

So,

 \because \rm \tan \theta = \frac{3}{2}

Therefore,

 \therefore \rm \frac{3}{2} = \frac{P}{4}

By Cross Multiplication,

 \implies \rm 2 \times P = 4 \times 3

 \implies \rm P = \frac{4 \times 3}{2}

 \implies \rm P = \frac{\cancel{4} \times 3}{\cancel{2}}

 \implies \rm P = 2 \times 3

 \implies \rm \bf P = 6 m

Hence, the height of the pole that casts that shadow is 6 m.

(b) To find :- The length of shadow cast by the another pole 12 m 25 cm high.

Solution :-

Length of the pole, P = 12.25 m

Since,

 \because \rm \tan \theta = \frac{P}{B}

Therefore,

 \therefore \rm \tan \theta = \frac{12.25}{B}

Also,

 \because \rm \tan \theta = \frac{3}{2}

So,

 \therefore \rm \frac{3}{2} = \frac{12.25}{B}

By Cross multiplication,

 \implies \rm 3 \times B = 2 \times 12.25

 \implies \rm B = \frac{2 \times 12.25}{3}

 \implies \rm B = \frac{24.50}{3}

 \implies \rm B = \cancel{\frac{24.50}{3}}

 \implies \rm \bf B = 8.16 m

Hence, the length of the shadow cast by that pole is 8.16 m or 8m 16cm.

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