Awesome q2a theme

0 votes

The area of triangle is calculated by the formula . If the angle A is measure with 1% error. Find the % error in area.

(A)

(B)

(C)

(D)

0 votes

Let S denote area,

S=$\frac{1}{2}bcsinA$

Given $\frac{\Delta A}{A}=0.01$

A=$\frac{\pi}{6} \implies \Delta A=0.01 \times \frac{\pi}{6} $

$dS=\frac{1}{2}d(bcsinA)=\frac{1}{2}(csinAdb+bsinAdc+bccosAdA)$

$dS=\frac{1}{2}d(bcsinA)=\frac{1}{2}(cdb+bdc+bccotAdA)$

$dS=\frac{1}{2}d(bcsinA)=\frac{1}{2}(\frac{db}{b}+\frac{dc}{c}+cotAdA)$

Now we have :

$\frac{dS}{S}=\frac{db}{b}+\frac{dc}{c}+\frac{dA}{tanA}$

$\implies \frac{ds}{S}=\frac{0.01 \times \frac{\pi}{6}}{tan30}=0.01 \times \sqrt 3 \times \frac{\pi}{6}$. To match an option maybe there is some approximation.

S=$\frac{1}{2}bcsinA$

Given $\frac{\Delta A}{A}=0.01$

A=$\frac{\pi}{6} \implies \Delta A=0.01 \times \frac{\pi}{6} $

$dS=\frac{1}{2}d(bcsinA)=\frac{1}{2}(csinAdb+bsinAdc+bccosAdA)$

$dS=\frac{1}{2}d(bcsinA)=\frac{1}{2}(cdb+bdc+bccotAdA)$

$dS=\frac{1}{2}d(bcsinA)=\frac{1}{2}(\frac{db}{b}+\frac{dc}{c}+cotAdA)$

Now we have :

$\frac{dS}{S}=\frac{db}{b}+\frac{dc}{c}+\frac{dA}{tanA}$

$\implies \frac{ds}{S}=\frac{0.01 \times \frac{\pi}{6}}{tan30}=0.01 \times \sqrt 3 \times \frac{\pi}{6}$. To match an option maybe there is some approximation.

9,003 questions

3,134 answers

14,387 comments

95,818 users