Subject:
MathAuthor:
deangelo23Created:
1 year agolet us consider the height of the tower 'h'
we know that 'h'= AC tan 30 degree = BC tan 30 degree
We know that AC=BC [∵ radius of circle]
Hence, we can say that △ABC is an isosceles triangle with AC=BC.
So, ∠ABC=∠BAC ...(2)
But ∠ACB=60
∘
[given]
⇒∠ABC+∠BAC+∠ACB=180
∘
⇒∠ABC+∠ABC+60
∘
=180
∘
[from (2)]
⇒2×∠ABC=120
∘
∴∠ABC=∠BAC=60
∘
Hence △ABC is an equilateral triangle.
∴AB=BC=CA=a ...(3)
∴h=atan30
∘
[form equation (1)]
=
3
a
Author:
esmeralda4hem
Rate an answer:
3Step-by-step explanation:
Solution
verified
Verified by Toppr
Correct option is C)
Let us consider the height of the tower as h
We know that, h=ACtan30
∘
=BCtan30
∘
...(1)
We know that AC=BC [∵ radius of circle]
Hence, we can say that △ABC is an isosceles triangle with AC=BC.
So, ∠ABC=∠BAC ...(2)
But ∠ACB=60
∘
[given]
⇒∠ABC+∠BAC+∠ACB=180
∘
⇒∠ABC+∠ABC+60
∘
=180
∘
[from (2)]
⇒2×∠ABC=120
∘
∴∠ABC=∠BAC=60
∘
Hence △ABC is an equilateral triangle.
∴AB=BC=CA=a ...(3)
∴h=atan30
∘
[form equation (1)]
=
3
a
Author:
jaydenb98a
Rate an answer:
1