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...
Answers
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
1/tan x = h/9
h = 9/tan x
tan x = 9/h ——(2)
From (1) and (2)
h/4 = 9/h
= 9*4
= 36
h =
h = 6
The height of Tower is 6m
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