Help with this math prob?

Question

The base of a right triangle is 8 cm and the angle is increasing at the constant rate of .03 radians per second, is the altitude h of the triangle increasing when h=13?

Answer 1

Ok, let's draw a right triangle ABC, the base : AC is 8 cm. Then, let's set a value to one side, side AB = a, and the altitude "h" :

And also, let's call angle A = theta

We need to find the rate of increase of the altitude :

Using trigonometry : a = 8*cos(theta)

then : h = 8*cos(theta)*sin(theta) = 4*sin(2theta)

You know the rate of the angle : d(theta)/dt = 0.03 rad/s

dh/dt = 4*d(sin2theta) / dt

dh / dt = 4*cos(2theta)2*d(theta)/dt

dh / dt = 8*cos(2theta)*0.03 rad/s

we can find the value of cos(2theta), is h = 13

13 = 8cos(theta*sin(theta) >>> impossible right ?

so well, the altitude is not increasing because there is no triangle then.

Note : I have seen the other procedures, and they are considering the altitude as one side of the triangle, and this is not always truth, be careful, but yes, if you consider the altitude as one side, then :

tan(theta) = h / 8

sec^2(theta)d(theta)/dt = 1/8*dh / dt

sec^2(theta) = tan^2(theta) + 1 = (h/8)^2 + 1 = (13/8)^2 + 1

3.64*0.03 = 1/8*dh/dt

dh/dt = 0.8736

Hope that helps

Answer 2

I think your question is AT WHAT RATE is the altitude increasing when h=13.

h/8 = tan Φ, so h = 8 tan Φ. Then
dh/dΦ = 8 sec² Φ, and since dΦ/dt is a constant 0.03 rad/s, by the chain rule,

dh/dt = dh/dΦ • dΦ/dt = 0.24 sec² Φ.

So when h = 13, what's sec² Φ? Using Pythagoras, cos Φ = 8/√(8²+13²), sec Φ = √(8²+13²) / 8, and sec² Φ = 233/64.

So when h=13, dh/dt = 0.24 • 233/64 = 0.87375 cm/s.

Answer 3

tan x =13/8=1,625

x=58 degrees=0,81 radians

0,81/0,03=27 seconds