
The angular depression of the top and foot of a tower as seen from the top of a second tower which is 150m high and standing on the same level as the first are α and β respectively. If \[tan\alpha {\text{ }} = \dfrac{4}{3}\]and \[tan\beta = \dfrac{5}{2}\], the distance between their top is:
A.100m
B.120m
C.110m
D.130m
Answer
542.7k+ views
Hint: To find the distance between the tops of the tower firstly. We have to draw a rough diagram. Since, we have given the angle of depression. This means, the height of the second tower is more than the height of the first tower. Angle α is made to the first tower and β at the bottom of the tower.
Complete step-by-step answer:
Now, ∠ FBD = ∠ BDE
This is because line BF is parallel to line DE and BD acts as transversal. The interior opposite angles formed by the transversal are equal.
∴ ∠ BDE = α
Similarly, ∠ FBC= ∠ BCA
Now, ABC formed is a triangle.
Let us consider that AC = y. here y is the distance between the towers. Also, height of tower 2 = AB = 150m.
Now in Δ ABC
$\tan \beta = \dfrac{{AB}}{{AC}} = \dfrac{{150}}{y}$
Also, we have given that $\tan \beta = \dfrac{5}{2}$
∴ $\dfrac{{150}}{y} = \dfrac{5}{2} \to y = \dfrac{{2 \times 150}}{5} = 60$
Y = 60m
Distance between the tower = 60m
Now in Δ BED, ∠ D=α and ∠ E = 90 degree
So, $\tan \alpha = \dfrac{{BE}}{{ED}}$. Also, we have given that $\tan \alpha = \dfrac{4}{3}$
Therefore $\dfrac{{BE}}{{ED}} = \dfrac{4}{3}$
Let us consider that BE = x
→ x is the height of tower 2 which is more than tower1.
Therefore $\dfrac{x}{{DE}} = \dfrac{4}{3}$
DE is also the distance between the two towers.
∴ DE= AC = y = 60m
Therefore $\dfrac{x}{{60}} = \dfrac{4}{3} \to x = \dfrac{{4 \times 60}}{3} = 80$
X= 80m
Now, distance between the two towers is given by BD.
∴ BDE is a right angled triangle.
Where BE is perpendicular, BD is hypotenuses and ED is base.
BY Pythagoras theorem,
${(hypotenuse)^2} = {(Base)^2} + {(perpendicular)^2}$
$\Rightarrow$ ${(BD)^2} = {(ED)^2} + {(BE)^2}$
$\Rightarrow$ ${(BD)^2} = {(y)^2} + {(x)^2}$
Putting values of x and y,
$\Rightarrow$ ${(BD)^2} = {(60)^2} + {(80)^2}$
$\Rightarrow$ ${(BD)^2} = 3600 + 6400$
$\Rightarrow$ ${(BD)^2} = 10000$
Take square root on both sides;
$
\Rightarrow (BD) = \sqrt {10000} \\
\Rightarrow BD = 100 \\
$
So, the distance between the top of the tower is 100m
Option (A) is correct.
Note: Angle of depression is a downward angle from the horizontal to the line of the sight from the observer to the same point of intersection.
Pythagoras theorem states that “ in a right angled triangle, the square of the hypotenuse side is equal to the sum of the squares of the other two sides of the triangle.
The tangent of an angle is equal to the side opposite to acute angle divided by the adjacent side of acute angle.
Complete step-by-step answer:
Now, ∠ FBD = ∠ BDE
This is because line BF is parallel to line DE and BD acts as transversal. The interior opposite angles formed by the transversal are equal.
∴ ∠ BDE = α
Similarly, ∠ FBC= ∠ BCA
Now, ABC formed is a triangle.
Let us consider that AC = y. here y is the distance between the towers. Also, height of tower 2 = AB = 150m.
Now in Δ ABC
$\tan \beta = \dfrac{{AB}}{{AC}} = \dfrac{{150}}{y}$
Also, we have given that $\tan \beta = \dfrac{5}{2}$
∴ $\dfrac{{150}}{y} = \dfrac{5}{2} \to y = \dfrac{{2 \times 150}}{5} = 60$
Y = 60m
Distance between the tower = 60m
Now in Δ BED, ∠ D=α and ∠ E = 90 degree
So, $\tan \alpha = \dfrac{{BE}}{{ED}}$. Also, we have given that $\tan \alpha = \dfrac{4}{3}$
Therefore $\dfrac{{BE}}{{ED}} = \dfrac{4}{3}$
Let us consider that BE = x
→ x is the height of tower 2 which is more than tower1.
Therefore $\dfrac{x}{{DE}} = \dfrac{4}{3}$
DE is also the distance between the two towers.
∴ DE= AC = y = 60m
Therefore $\dfrac{x}{{60}} = \dfrac{4}{3} \to x = \dfrac{{4 \times 60}}{3} = 80$
X= 80m
Now, distance between the two towers is given by BD.
∴ BDE is a right angled triangle.
Where BE is perpendicular, BD is hypotenuses and ED is base.
BY Pythagoras theorem,
${(hypotenuse)^2} = {(Base)^2} + {(perpendicular)^2}$
$\Rightarrow$ ${(BD)^2} = {(ED)^2} + {(BE)^2}$
$\Rightarrow$ ${(BD)^2} = {(y)^2} + {(x)^2}$
Putting values of x and y,
$\Rightarrow$ ${(BD)^2} = {(60)^2} + {(80)^2}$
$\Rightarrow$ ${(BD)^2} = 3600 + 6400$
$\Rightarrow$ ${(BD)^2} = 10000$
Take square root on both sides;
$
\Rightarrow (BD) = \sqrt {10000} \\
\Rightarrow BD = 100 \\
$
So, the distance between the top of the tower is 100m
Option (A) is correct.
Note: Angle of depression is a downward angle from the horizontal to the line of the sight from the observer to the same point of intersection.
Pythagoras theorem states that “ in a right angled triangle, the square of the hypotenuse side is equal to the sum of the squares of the other two sides of the triangle.
The tangent of an angle is equal to the side opposite to acute angle divided by the adjacent side of acute angle.
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