ABC is a right angled triangle on a horizontal ground. AD is a vertical tower. BAC is 90 degrees, ACB is 35 degrees and ABD is 52 degrees. C is 66 metres. Find the height of the tower and the angle of elevation of the top of the tower from C?
Sounds like some pretty straightforward calculus. Any idea how to get started with this?
C is an angle. Which side is 66? CB or CA?
In any case, you can use sin or cos to find the sides of ABC
Then the height AD can be found via
AD/AB = tan52°
tan(ACD) = AD/AC
To solve this problem, we can use the properties of right-angled triangles and trigonometric ratios.
First, let's find the height of the tower (AD).
Since triangle ABC is a right-angled triangle, we can use the tangent ratio:
tangent(ADB) = height of the tower (AD) / distance from A to the point directly below the tower (AB)
Given that ∠ADB is 52 degrees, we can substitute the values:
tangent(52) = AD / AB
Now, let's find the value of AB:
In triangle ABC, we know that C is 66 meters. And, we have the angle ACB as 35 degrees. We can use the sine ratio:
sine(35) = height of the tower (AD) / hypotenuse (AC)
Now, substitute the values:
sine(35) = AD / 66
You can rearrange this equation to find AD:
AD = 66 * sine(35)
Now, we can use the value of AD and AB to calculate the value of AB:
tangent(52) = AD / AB
Rearrange the equation to find AB:
AB = AD / tangent(52)
Now, you have the values of AD and AB, which can be used to find the height of the tower.
To find the angle of elevation of the top of the tower from C, we can use the inverse tangent function:
angle of elevation = inverse tangent(AD / BC)
Substitute the values:
angle of elevation = inverse tangent(AD / 66)
Calculate this value, and you will have the angle of elevation of the top of the tower from C.