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?

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- Aug 29th 2008, 09:45 AMmatthewspjTrigonometric 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?

- Aug 29th 2008, 10:49 AMJhevon
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