Two towers face each other separated by a distance d = 15 m As seen from the top of - Maths - Some Applications of Trigonometry

Question

Two towers face each other separated by a distance d = 15 m As
seen from the top of the first tower, the angle of depression of
the second tower's base is 600 and that of the top is
300 What is the height (in meters) of the second
tower?

Manbar Singh · Accepted Answer

Let AB be the first tower and CD be the second tower Let BD be the
distance between two towers Now BD = 15 mDraw CL⊥BC Now, CL =
BD = 15 mLet ∠ACL = 30° and ∠ADB = 60° Now, in
∆ABD,tan 60° = ABBD⇒3 = AB15⇒AB = 153 m Now, in
∆ALC, tan 30° = ALLC⇒13 = AL15⇒AL = 153= 1533 =
53 mNow, LB = AB - AL = 153 - 53 = 103m = 17 32 mNow, CD = LB = 17
32 mSo, height of second tower = CD = 17 32 m

Two towers face each other separated by a distance d = 15 m. As seen from the top of the first tower, the angle of depression of the second tower's base is 600 and that of the top is 300. What is the height (in meters) of the second tower?

Two towers face each other separated by a distance d = 15 m. As seen from the top of the first tower, the angle of depression of the second tower's base is 60⁰ and that of the top is 30⁰. What is the height (in meters) of the second tower?