Two points A and B are on the same side of a tower and in the same straight line with its base. The angle of depression of these points from top of the tower are $60 °$ and $45 °$ respectively. If the height of the tower is 15 m, then the distance between these points is ____.

Question

Two points A and B are on the same side of a tower and in the same straight line with its base. The angle of depression of these points from top of the tower are $60 °$ and $45 °$ respectively. If the height of the tower is 15 m, then the distance between these points is ____.

Infinity Learn · Accepted Answer

We must calculate the distance between points 'A' and 'B.' Assume that the distance between points 'A' and 'B' from the base of the tower is x.m and 'y' m, respectively.
We can see from the diagram that-
Distance between points ‘A’ and ‘B’ =

(y - x) m

…………………………….. (1)
To find the distance between ‘A’ and ‘B’, we need to find ‘x’ and ‘y’.
Let us find ‘x’ first-
In the diagram, we can see that-
∠NMA+∠AMO=

90 °

(Complementary angles)
Given angle of depression for point A =

60 °

.
⇒∠NMA=

60 °

So,

60 °

+∠AMO=

90 °

⇒∠AMO=

90 ° - 60 °

⇒∠AMO=

30 °

We know that tan

θ = \frac{perpendicular}{base}

In ΔOMA -
For ∠AMO -
Perpendicular =OA=

xm

Base =OM=15m
So, tan∠AMO=

\frac{OA}{OM}

.
On putting ∠AMO=

30 °

, OA=

xm

and OM=15m, we will get-
tan

30 ° = \frac{x}{15}

We know tan

30 ° = \frac{1}{\sqrt 3}

On putting tan

30 ° = \frac{1}{\sqrt{3}}

, we will get-
⇒.

\frac{1}{\sqrt{3}} = \frac{x}{15}

On multiplying both sides by 15, we will get-
⇒

\frac{15}{\sqrt{3}} = x

On multiplying both numerator and denominator by √3 in LHS, we will get
⇒

\frac{15 \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = x

⇒

\frac{15 \sqrt{3}}{3} = x

⇒

5 \sqrt{3} = x

⇒

x =

5

\sqrt{3}

m
Now let us find ‘y’
In the diagram, we can see that ∠NMB+∠BMO=

90 °

(Complementary angles)
Given angle of depression for point B=

45 °

⇒∠NMB=

45 °

So,

45 °

+∠BMO=

90 °

∠BMO=

90 ° - 45 °

∠BMO=

45 °

Again, we will apply tan

θ = \frac{perpendicular}{base}

for ∠BMO in triangle OMB.
In ΔOMB -
Fro ∠BMO -
Perpendicular =OB=y.m
Base =OM=

x

m
So, tan ∠BMO=

\frac{OB}{OM}

On putting ∠BMO=

45 °

, OB=y.m and OM=15m, we will get-
tan

45 ° = \frac{ym}{15 m}

.
We know that tan

45 ° = 1

.
On putting tan

45 ° = 1

, we will get-
1=

\frac{y}{15}

On multiplying both sides by 15, we will get-
15=y
⇒y=15m
Now, we have obtained

x = 5 \sqrt 3

m and y=15m, So, putting

x = 5 \sqrt 3

m and y=15m , we will get-
Distance between A and B =15−5(1.732)
=15−8.660
=6.34
Hence, the required distance between points A and B is 6.34m.

Two points A and B are on the same side of a tower and in the same straight line with its base. The angle of depression of these points from top of the tower are $60 °$ and $45 °$ respectively. If the height of the tower is 15 m, then the distance between these points is ____.

Detailed Solution

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Two points A and B are on the same side of a tower and in the same straight line with its base. The angle of depression of these points from top of the tower are 60°and 45° respectively. If the height of the tower is 15 m, then the distance between these points is ____.

Detailed Solution

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Get Expert Academic Guidance – Connect with a Counselor Today!

Two points A and B are on the same side of a tower and in the same straight line with its base. The angle of depression of these points from top of the tower are $60 °$ and $45 °$ respectively. If the height of the tower is 15 m, then the distance between these points is ____.