A 6m 40cm high pole casts a shadow casts a shadow 4m 80 cm long . find:
i the length of shadow cast by another pole 12m 80cm high .
ii, the height of a pole which casts a shadow 2m 40 cm
pls answer
Answers
Answer:
Given :-
Height of the pole (P) = 6m 40 cm
Length of the shadow (B) = 4m 80 cm
Solution :-
In Right angled Δ ABC formed,
Perpendicular, P = 6.40 m
Base, B = 4.80 m
Since,
\because \rm \tan \theta = \frac{P}{B}∵tanθ=
B
P
Therefore,
\therefore \mathrm{\tan \theta = \frac{6.40}{4.80}}∴tanθ=
4.80
6.40
\implies \mathrm{\tan \theta = \cancel{\frac{6.40}{4.80}}}⟹tanθ=
4.80
6.40
\implies \boxed{\bold{\mathrm{\tan \theta = \frac{4}{3}}}}⟹
tanθ=
3
4
(a) To find :- The length of shadow cast by the another pole 12 m 80 cm high.
Solution :-
Length of the pole, P = 12.80 m
Since,
\because \rm \tan \theta = \frac{P}{B}∵tanθ=
B
P
Therefore,
\therefore \rm \tan \theta = \frac{12.80}{B}∴tanθ=
B
12.80
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., \tan \thetatanθ is equal in both poles.
So, \tan \theta = \frac{4}{3}tanθ=
3
4
Now,
\therefore \rm \frac{4}{3} = \frac{12.80}{B}∴
3
4
=
B
12.80
By Cross multiplication,
\implies \rm 4 \times B = 3 \times 12.80⟹4×B=3×12.80
\implies \rm B = \frac{3 \times 12.80}{4}⟹B=
4
3×12.80
\implies \rm B = \frac{3 \times \cancel{12.80}}{\cancel{4}}⟹B=
4
3×
12.80
\implies \rm B = 3 \times 3.2⟹B=3×3.2
\implies \rm \bf B = 9.6 m⟹B=9.6m
Hence, the length of the shadow cast by that pole is 9.6 m or 9m 60cm.
(b) To find :- The height of a pole which casts a shadow 2m 40cm.
Solution:-
Length of the shadow, B (Base) = 2m 40 cm = 2.4 m
Since,
\because \rm \tan \theta = \frac{P}{B}∵tanθ=
B
P
Therefore,
\therefore \rm \tan \theta = \frac{P}{2.4}∴tanθ=
2.4
P
Also,
\because \rm \tan \theta = \frac{4}{3}∵tanθ=
3
4
So,
\therefore \rm \frac{4}{3} = \frac{P}{2.4}∴
3
4
=
2.4
P
By Cross Multiplication,
\implies \rm 3 \times P = 4 \times 2.4⟹3×P=4×2.4
\implies \rm P = \frac{4 \times 2.4}{3}⟹P=
3
4×2.4
\implies \rm P = \frac{4 \times \cancel{2.4}}{\cancel{3}}⟹P=
3
4×
2.4
\implies \rm P = 4 \times 0.8⟹P=4×0.8
\implies \rm \bf P = 3.2 m⟹P=3.2m
Hence, the height of the pole that casts that shadow is 3.2 m or 3 m 20 cm.