Question

A vertical tower stands on a horizontal plane and is surmounted by a flag-staff of height 7m. From a point on the plane, the angle of elevation of the bottom of the flag-staff is 30° and that of the top of the flag-staff is 45°. Find the height of the tower.

Answer 1

Answer:

9.56 m

Step-by-step explanation:

From figure:

Let BC be the height 'h'.

∠ADC = 45° and ∠BDC = 30°.

(i) From ∠BCD:

⇒ BC/DC = tan 30°

⇒ h/DC = (1/√3)

⇒ DC = h√3

(ii) From ∠ACD:

AC/DC = tan 45°

⇒ (AB + BC)/DC = 1

⇒ (7 + h)/DC = 1

⇒ 7 + h = DC

From (i) & (ii), we get

⇒ h√3 = 7 + h

⇒ h√3 - h = 7

⇒ h(√3 - 1) = 7

⇒ h = (7/√3 - 1)

⇒ h = (7/√3 - 1) * (√3 + 1)/(√3 + 1)

⇒ h = 7(√3 + 1)/(3)² - 1

⇒ h = 7(√3 + 1)/2

⇒ h = 7(1.73 + 1)/2

⇒ h = 7(2.73)/2

⇒ h = 19.11/2

⇒ h = 9.5 m

Therefore, height of the tower = 9.56 m.

Hope it helps!

Answer 2

$\huge\mathfrak\red{Answer :}$


According to question we have given;

AB = 7

BC = h

/_BCD = 30°

/_ADC = 45°

Let DC = x

BC = h = ?

* Figure is in Attachment !!

In ∆ACD

$\dfrac{AC}{DC}$ = tan 45°

$\dfrac{(7\:+\:h)}{x}$ = tan 45°

$\dfrac{(7\:+\:h)}{x}$ = 1

As tan 45° = 1

$7\:+\:h\:=\:x$ ...... (1)

In ∆BCD

$\dfrac{BC}{DC}$ = tan 30°

$\dfrac{h}{x}$ = tan 30°

$\dfrac{h}{x}$ = $\dfrac{1}{\sqrt{3}}$

$h\sqrt{3}\:=\:x$ ...... (2)

Put value of x in (1)

$7\:+\:h\:=\:h\sqrt{3}$

$7 \: = \: h( \sqrt{3} \: - \: 1)$

$h \: = \: \frac{7}{( \sqrt{3} \: - \: 1)}$

$h \: = \: \frac{7}{( \sqrt{3} \: - \: 1) } \: \times \: \frac{ (\sqrt{3} \: + \: 1)}{ (\sqrt{3} \: + \: 1) }$

$h \: = \: \frac{ 7(\sqrt{3} \: + \: 1) }{ { (\sqrt{3}) }^{2} \: - \: {(1)}^{2} }$

$h \: = \: \frac{7(1.732 \: + \: 1)}{2}$

$h \: = \: \frac{7(2.732)}{2}$

$h \: = \: \frac{19.24}{2}$

$h \: = \: 9.562 \: m$

=====================================