Question: A 40 m tall building and a tower are standing in front of each other. Angles of elevation to the top of the tower from the top and base of the building are 45o and 60o, respectively. How do I calculate the height of the tower? Please help me to calculate

Posted by: Syed G. on 25.08.2017

You can use simple trigonometry ratio. We know that [email protected]= P/B which give height of tower as 94.64m

tan 45= H/B

B=H

tan 60= H+40/H

1.732H-H=40

.732H=40

H=54.6

​tan 60 = (40+y)/ Z      let y be height building level to top of tower , z be distance

Height of the tower = 40 + 40/(sqrt(3)-1) = 20 (3+sqrt(3))
Let height of tower be x, and distance between the two buildings be d. From top of building we get from right triangle, (x-40)/d = tan 60 and from base x/d = tan 45 x-40 = d and x-40/(x) = 0.866 x-40 = 0.866 x x= 298.51 m.
First of all draw two parallel lines in vertical direction. One of them should be taller than the other. Name the smaller line as building or b whichever is convenient to you. And name tower for the taller line. Tower is taller than building. Why ? It is obvious from the problem as the angle of elevation to the top of the tower from the top and base of the building is 45° and 60°. Back to the diagram. Connect the top and bottom of the building to top of the tower. Draw two parallel horizontal lines from top and base of the building till it intersect with line of tower. Half work is done. Now mention the angle of elevation in the diagram. Now we have the height of the building. We have diagram of above problem as rectangle on which triangle is mounted. Horizontal line from the top of the building which intersect tower at certain point say 'p'. Now height of tower upto point p is 40 mtr. As sides of rectangles are equal. Now suppose rest of the height of the tower is x. Now we have 2 different equtions. ie 1.tan45=x/base 2.tan60=(40+x)/base. So, Base=x/tan45 Base=x/tan60 x/tan45 = x/tan60 solve for x this implies x=54.6448 m So hight of tower is x+40 implies height of tower is 94.64 mtr.
94.57 as ques tells tower & building are opposite to each other so draw sketch showing all detail after apply trignometary u will get your ans
First draw the sketch as... Building and Tower facing opposite side and parallel where the height of tower is more than building. draw a line from top of building parallel to the road(base) Connect top of the building and top of tower, form 45 degree. Connect bottom of the building and top of tower, form 60 degree. Let the distance between tower and building be Y and Height of tower be 40+X(extra height of tower exceeding building). Now, Tan 45= X/Y=1.. => X=Y. Tan 60= (40+X)/Y = Root 3... => X= 40/0.732= 54.64.
Let the tower be X m more than the building .then since elevation of the tower from the top of building is 45 degree so it will form an isosceles triangle .so the distance between the two will also be Xm. now using angle 60 degree triangle tan60=X+40/X. solving this X=20*(root 3+1) so height of tower =40+X=40+20*(root3+1)=94.64m
We first draw the 40m building as a straight line and the tower (taller than the building)at a distance y from the building (the lines should be parallel to each other) then from the top of the building a 45° to the top of the tower and from the base of the building line a 60°to the top. The tower is x meters taller than the building. Now we'd use trigonometry, since the lines are parallel, the base of the 45° and 60° triangle is same. Using tan 45 and tan 60, equate both of them using the common base. So you have, (40+×)÷tan 60 = ×÷ tan 45. Solving for x, you'd get ×= 54.64. We've considered the tower to be x meters taller, hence add that to 40m, so we get 40+ 54.64= 94.64m
First we have to draw the graph according to question, Tower is taller than building and both are parallel to each other. We calculate then we find answers. 20(root3+1)
Let the tower be X m more than the building .then since elevation of the tower from the top of building is 45 degree so it will form an isosceles triangle .so the distance between the two will also be Xm. now using angle 60 degree triangle tan60=X+40/X. solving this X=20*(root 3+1) so height of tower =40+X=40+20*(root3+1)=94.64m
Heght be x , thn tan45 = (x-40)/diff. between the feet of building & tower diff = x - 40 also tan 60= x/x-40 root3 *(x-40) = x ie. x = 94.6
Let, height of the tower be x metre. So, tan 45 = (x-40)/(diff. between the feet of building & tower). Thus, diff. between their feet = (x-40) metre [since, tan 45 =1]. Again, tan 60 = x/(x-40) . Or, rt3*(x-40) = x . Or, x(rt3 - 1) = 40*rt3 . Or, x*0.732 = 40*1.732 . Thus, x = 94.64 (approx) . Hence, height of the tower is 94.64 m Ans.
Syed, this type of problems can be solved by making a diagram. Diagram can not be pasted here, share ur email, i will send u daigram. Let assume x be height of tower, b is distance between tower and building. Angle of elevation from base is 60 degree Angle of elevation from top of building is 45 degree. So in right angle triangle using perpendicular and base formula for tan. Equation 1 : tan 60 = x/b Equation 2 : tan 45 = (x-40)/b Dividing both equation we get : tan 60 / tan 45 = x/(x-40) Substtuting values and solving above equation we get x = 94.64 ,meters which is height of the tower. Hope it helps. For more queries, contact me at [email protected]
Let the height of tower be h metre more than that of building and distance between tower and building be x metre then tan 45 = h/x, therefore h = x. now tan 60 = (40+h)/x, substitute x by h and calculate the value of h from tan 60 = (40+h)/h, h = 54.64 metres and height of tower is 54.64+40 = 94.64 metres
Let the tower be X m more than the building .then since elevation of the tower from the top of building is 45 degree so it will form an isosceles triangle .so the distance between the two will also be Xm. now using angle 60 degree triangle tan60=X+40/X. solving this X=20*(root 3+1) so height of tower =40+X=40+20*(root3+1)=94.64m

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