Question

From the top of a 60 m high building, the angles of depression of the top and the bottom of a tower are 45 and 60 respectively. Find the height of the tower. [ Take 3 = 1.73 ]

Answer 1

Answer:

Hence, the height of the tower is:

25.4 m

Step-by-step explanation:

From the figure we could see that we obtain 2 right angled triangle as:

ΔAED and ΔABC

Hence, we will apply the trignometric identity in both the triangles as:

In ΔAED

$\tan 45\degree=\dfrac{x}{l}\\\\1=\dfrac{x}{l}\\\\x=l$

Also, in ΔABC we have:

$\tan 60=\dfrac{60}{l}\\\\\sqrt{3}=\dfrac{60}{l}\\\\l=\dfrac{60}{\sqrt{3}}\\\\l=20\sqrt{3}\text{m}$

Hence,

$x=20\sqrt{3}\text{m}$

Hence, the height of the tower is:

$60-x=60-20\sqrt{3}\\\\=60-20\times 1.73\\\\=60-34.6\\\\=25.4\text{m}$

Hence, the height of the tower is:

25.4 m

Answer 2

Answer:

Answer:

Hence, the height of the tower is:

25.4 m

Step-by-step explanation:

From the figure we could see that we obtain 2 right angled triangle as:

ΔAED and ΔABC

Hence, we will apply the trignometric identity in both the triangles as:

In ΔAED

\begin{gathered}\tan 45\degree=\dfrac{x}{l}\\\\1=\dfrac{x}{l}\\\\x=l\end{gathered}

tan45°=

l

x

1=

l

x

x=l

Also, in ΔABC we have:

\begin{gathered}\tan 60=\dfrac{60}{l}\\\\\sqrt{3}=\dfrac{60}{l}\\\\l=\dfrac{60}{\sqrt{3}}\\\\l=20\sqrt{3}\text{m}\end{gathered}

tan60=

l

60

3

=

l

60

l=

3

60

l=20

3

m

Hence,

x=20\sqrt{3}\text{m}x=20

3

m

Hence, the height of the tower is:

\begin{gathered}60-x=60-20\sqrt{3}\\\\=60-20\times 1.73\\\\=60-34.6\\\\=25.4\text{m}\end{gathered}

60−x=60−20

3

=60−20×1.73

=60−34.6

=25.4m

Hence, the height of the tower is:

25.4 m