Question

Trigonometry 11.

1. is the angle of elevation of the top of the tower from two points at a distance of 4m and 9M from the base of the tower and in the same straight line with it are complementary find the height of the tower...

Answer 1

Answer:

Step-by-step explanation:

Let the height of the tower ,AB be ‘h’

Let C be a point which is 4m away from the tower

I.e BC = 4m

Angle of elevation from point C be ‘x’

Angle C and Angle D are complementary angles .

So, Angle C + Angle D = 90°

Angle D = 90°-angle C

=> 90-x

D is a point 9m away from the tower I.e BD= 9m

In right triangle,ABC ,Angle B = 90°

AB =h ,BC=4m ,Angle ACB=x

Using Trigonometric ratios,

Tan C = AB/BC

Tan x = h/4—(1)

In right triangle ABD,Angle B = 90°

AB = h,BD=9m,Angle D = 90-x

Using Trigonometric ratios,

Tan D = AB/BD

Tan (90-x) = h/9

Cot x = h/9 $\boxed[tan(90-A)=cotA$

1/tan x = h/9

h = 9/tan x

tan x = 9/h ——(2)

From (1) and (2)

h/4 = 9/h

$h^{2}$= 9*4

$h^{2}$= 36

h = $\sqrt{36}$

h = 6

The height of Tower is 6m

Answer 2

Answer:

6 m

Step-by-step explanation:

Let the angles be α & β .

Given :

α + β = 90°

So, β = 90° - α

tan β = tan(90° - α) = cot α .....(1)

From figure,

tan α = h/4 &

tan β = h/9 [from (1)]

So,

(tan α )×(tanβ ) = (h/4)*(h/9)

(tan α )×(cot α) = (h/4)*(h/9)

1 = h²/36

h = ± 6

Since, -6 is not possble .

So, Height of tower = 6 m