Question

Trigonometry. Applications Question


From the top of a 7 m high building, the angle of elevation of a cable tower is
60 degree and the angke of depression of its foot is 45 degree. Determine the height of the tower.


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Answer 1

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The height of cable tower will be 19.124 m.

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Let the building as DC where DC is 7m and angle of elevation and depression to the top and foot of the cable tower is 60° and 45° respectively .

Now let the tower as AB . Here EB is equivalent to DC and AE is the additional height of tower. Angle C and D is 90° .

→ Now in ∆ BCD

$\frac{dc}{bc} = \tan45$

Because we know that tan theta = perpendicular/ base

→ Putting value of DC and tan 45°

$\frac{7}{x} = 1$

$x(cb) \: = 7m$

Now as we know that BC = DE

→ Now in ∆ AED , Again applying tan

$\frac{ae}{ed} = \tan(60)$

replacing ED by BC

$\frac{ae}{7} = \sqrt{3}$

So, AE = 7√3 .

Now total height of tower is AE + EB

Total height = 7 + 7√3

AB. = 7(1+1.732) { √3 = 1.732}

AB. = 19.124 m

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Answer 2

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❤❤Hope this will helpful for you❤❤