Question

A vertical tower stands on a horizontal plane and is s surmounted by a flagstaff of height 12m. At a point on the plane , the angle of elevation of the bottom of the flagstaff is 45° and of the top of the flagstaff is 60°. determine the height of the tower and the horizontal distance. (write the answer in to 2 decimal places).​

Answer 1

Answer:

Let the height of the tower be h.

Hence total height of the structure is h+12m.

Let the distance of the point from the structure be x.

Now

tan45

0

=1=

x

h

Hence

h=x

And

tan60

0

=

x

h+12

=

3

Hence

h+12=

3

x

h+12=

3

h

h=

3

−1

12

=16.392m

Step-by-step explanation:

hope it helps

Answer 2

        $\text{The question has been visualized in the attachment below}$

        $\text{According to the question:}$

        $BC = 12\:m$

        $\angle{CDA} = 45^{\circ}$

        $\angle{BDA} = 60^{\circ}$

        $\text{Let }AC = h \text{ and }AD = x$

$\implies \text{In }\triangle{ACD}$

        $\angle{CDA} = 45^{\circ}$

$\implies \: tan\angle{CDA} = tan45$

$\implies \: \dfrac{h}{x} = 1$

$\implies \: x = h$

        $\text{In }\triangle{ABD}$

        $\angle{BDA} = 60^{\circ}$

$\implies \: tan\angle{BDA} = tan60$

$\implies \: \dfrac{12 + h}{x} = \sqrt3$

$\implies \: 12 + h = \sqrt3x$

        $\text{We proved that }x = h$

$\implies \: 12 + x = \sqrt3x$

$\implies 12 = (\sqrt3 - 1)x$

$\implies \dfrac{12}{\sqrt3 - 1} = x$

        $\text{We know that }\sqrt3 = 1.73$

$\implies \: \dfrac{12}{0.73} = x$

$\implies \: \boxed{x = h \approx 16.43\:m}$