# Calculus - 13

The area of triangle is calculated by the formula $\frac{1}{2} bc\text{ }sin A$ . If the angle A is measure $30^\circ$ with 1% error. Find the % error in area.

(A) $\frac{\pi}{\sqrt{3}}$

(B) $\frac{\pi}{2\sqrt{3}}$

(C) $\frac{2\pi}{\sqrt{3}}$

(D) $\frac{\pi}{2}$

asked May 30, 2019
reshown May 31, 2019

Pi / 2 . Sqrt 3

1/2 bc sin A  differentiating

1/2 bc cos A dA
answered May 31, 2019 by (880 points)
edited May 31, 2019
where pie comes ??
dA is error just like d theta ...So substitute it...
please show me how u did it
1/2 bc sin A
differentiating....
1/2 bc cos A dA*
here A = 30 degrees... A* = 1/(pie/6) .=  6/pie  (1 percent)
1/2 * sqrt 3 /  2 * 6/pie
(3*sqrt 3) / 2*pie
21 * sqrt 3 / 44 which is equal to 0.85

now come to the option pie / 2* sqrt 3
22/14*sqrt 3 approximately equal to that .....
Thats y my answer is that.

Note: Remaining options are not near to that...
I am sorry to ask it late but how did u write "here A = 30 degrees... A* = 1/(pie/6) .=  6/pie  (1 percent)"
kindly elaborate
Also in 1/2 bc sin A
differentiating....
1/2 bc cos A dA* ( I think it should be dA, isn't it,if not why so ???))
Hii

here 1% in 30 degrees is 1/30 degrees ... .dA=6/pie
1/2 bc are constants... differentiate sinA...... cosA dA  = cos 30  * 6/pie
then follows.....
this is not a generalised method. i just used to solve that particular problem.
If I get any other method during revison time. I will update u.
Shouldn't be theta=pie/6 or A=pie/6,then how u wrote dA=6/pie ??

Also please answer to this "1/2 bc cos A dA* ( I think it should be dA, isn't it,if not why so ???))"
answered May 31, 2019 by (3,750 points)
please sir provide a solution..
answered May 31, 2019 by (880 points)
My method:

1/2 bc sin A
differentiating....
1/2 bc cos A dA*
here A = 30 degrees... A* = 1/(pie/6) .=  6/pie  (1 percent)
1/2 * sqrt 3 /  2 * 6/pie
(3*sqrt 3) / 2*pie
21 * sqrt 3 / 44 which is equal to 0.85

now come to the option pie / 2* sqrt 3
22/14*sqrt 3 approximately equal to that .....
Thats y my answer is that.

Note: Remaining options are not near to that...
$\\ a=1/2*bc*sinA\\ \\ \frac{\partial a}{\partial A}=1/2*bc*cosA\\ \\ \frac{\partial a}{\partial A}=1/2*bc*sinA*(cosA/sinA)\\ \\ \frac{\partial a }a= \frac{\partial A}\ *cosA/sinA\\ \\ \frac{\partial a }a=\frac{\partial A}A*cotA*A\\ \\ \frac{\partial a}a= 1*cot(30)*\pi /6\\ \\ =\sqrt{3}/6*\pi \\ \\ =\pi /2\sqrt{3}$