Thread: Trigonometric differentiation problem

1.The base of an isosceles triangle is 20cm and the altitude is increasing at the rate of 1cm/min. At what rate is the base angle increasing when the area is 100cm^2?

Originally Posted by matthewspj

1.The base of an isosceles triangle is 20cm and the altitude is increasing at the rate of 1cm/min. At what rate is the base angle increasing when the area is 100cm^2?

A is the area, h is the height, b is the length of the base and $\displaystyle \theta$ is the base angle.

recall that $\displaystyle A = \frac 12 bh$

this means, when A = 100, h = 10.

we are also told that $\displaystyle \frac {dh}{dt} = 1$. b is constant at 20cm, and so $\displaystyle \frac {db}{dt} = 0$, but we wont even have to worry about that.

draw your triangle, and bisect the base with a vertical line going through the top vertice. $\displaystyle \theta$ is the base angle and we form a right triangle with the opposite side to theta being h and the adjacent side being 10.

Thus we have $\displaystyle \sin \theta = \frac h{10}$

now differentiate implicitly and continue