Question

A man 1.7 m tall standing at the foot of a tower sees the top of a building 50 m away at an angle of elevation 50. find the height of the tower.find height of the building. ​

Answer 1

Answer:

kya mtlb bhaya mne smj na aaya

Answer 2

Your Question: A man, $1.7$ metres tall, standing at the foot of a tower sees the top of a building $50$ metres away at an angle of elevation $60$°. On climbing the top of the tower, he sees it at an angle of elevation of  $50$°. Draw a rough figure according to the given data. Compute the height of the tower and the building.

• (sin $50$° $=0.77$
• cos $50$° $=0.64$
• tan $50$° $=1.19$ 
• And
• sin $60$° $=0.87$
• cos $60$° $=0.50$
• tan $60$° $=1.73$ )

Basic Concept[s] - •} Angle of Elevation : The angle of elevation of the point viewed is the angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level, i.e., The case when we raise our head to look at the object.

(Refer Figure $1$ for more details about angle of elevation)

(Refer Figure $2$ for details about right angled triangle)

Solution:

Come, let's draw the figure now.

(For the Exact figure, refer figure number $3$)

In Figure number $3$:

• DF is the height of man and it is $1.7$ metres.
• CF is the height of the tower
• AG is the height of the building

If we will observe the figure carefully, we will find that:

• CF = BG
• DF = EG = $1.7$ metres
• DE = FG = CB = $50$ metres (Given)

Let's observe Triangle ADE carefully. After observing Triangle ADE carefully, we will find that:

• The angle made is $60$°.
• We know that tan and cot are the only trigonometric ratios which deal with Perpendicular and Base of a right - angled triangle. Let's take the trigonometric ratio of tan (Tangent).
• We also know that  $tan\theta = \frac{Perpendicular}{Base}$.

• Here, the angle made is $60$°. So, our theta is $60$° i.e $\theta=60$°. Putting   $60$° in $tan\theta = \frac{Perpendicular}{Base}$, we get:

$tan60$° $= \frac{Perpendicular}{Base}$

• We know that $tan60$° $=\sqrt{3}$. Putting  $tan60$° $=\sqrt{3}$ in $tan60$° $= \frac{Perpendicular}{Base}$ , we get:

$\sqrt{3} = \frac{Perpendicular}{Base}$

• We know that in Triangle ADE, perpendicular (AE) is missing and base (DE) is   $50$ metres.

$\sqrt{3} = \frac{AE}{DE} = \frac{AE}{50}$

• Take $50$ to the other side and change its operation to multiply.

$\sqrt{3} \times 50 = AE$

• We know that $\sqrt{3} = 1.73$. Putting $\sqrt{3} = 1.73$ in $\sqrt{3} \times 50 = AE$, we get:

$1.73 \times 50 = AE$

• Multiply accordingly.

$AE = 86.5$ metres

• From figure no $3$, we can observe that $AG = AE + EG$.
• We know that $AE = 86.5$ metres and $EG = 1.7$ metres.
• Putting $AE = 86.5$ metres and $EG = 1.7$ metres in $AG = AE + EG$, we get:

$AG = 86.5 + 1.7$ metres

• Add accordingly:

$AG = 88.2$ metres

So, the height of the building is $88.2$  metres.

Now, let's observe Triangle ABC carefully. After observing Triangle ABC carefully, we get:

The angle made is $50$°.

We know that tan and cot are the only trigonometric ratios which deal with Perpendicular and Base of a right - angled triangle. Let's take the trigonometric ratio of tan (Tangent).We also know that  $tan\theta = \frac{Perpendicular}{Base}$.

• Here, the angle made is $50$°. So, our theta is  $50$° i.e $\theta=50$°. Putting  $\theta=50$° in $tan\theta = \frac{Perpendicular}{Base}$, we get:

$tan50$° $= \frac{Perpendicular}{Base}$

• We are given in the question that $tan50$° $= 1.19$. Putting  $tan50$° $= 1.19$ in  $tan50$° $= \frac{Perpendicular}{Base}$ , we get:

$1.19 = \frac{Perpendicular}{Base}$

• We know that in triangle ABC, perpendicular (AB) is missing and base (BC) is   $50$ metres.

$1.19 = \frac{AB}{BC} = \frac{AB}{50}$

• Take $50$ to the other side and change its operation to multiply.

$1.19 \times 50 = AB$

• Multiply accordingly.

$AB = 59.5$ metres

• From figure no $3$, we can observe that $BE = CD = AE - AB$
• We know that $AE = 86.5$ metres and $AB = 59.5$ metres.
• Putting  $AE = 86.5$ metres and $AB = 59.5$ metres in $BE = CD = AE - AB$, we get:

$BE = CD = 86.5 - 59.5$ metres

• Subtract accordingly:

$BE = CD = 27$ metres

So, the height of the tower is  $27$ metres.

Final Answer:

• Hence, the height of the building is $88.2$  metres and the height of the tower is  $27$ metres.

Hope it Helps!

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