The area of triangle is measured in terms of b,c,A. If A=63∘ and there is an error of 15' in A; the percentage error in the area is

Ar=12bcsin⁡A,A=63∘,ΔA=15'⇒ΔAr=1560×π180⇒ΔArAr×100=cot⁡A.ΔA×100=5π36cot⁡63∘

The area of triangle is measured in terms of b,c,A. If A=63∘ and there is an error of 15' in A; the percentage error in the area is

Ar=12bcsin⁡A,A=63∘,ΔA=15'⇒ΔAr=1560×π180⇒ΔArAr×100=cot⁡A.ΔA×100=5π36cot⁡63∘

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$The area of triangle is measured in terms of b, c, A . If A = 63^{\circ} and there is an error of 15' in$ $A; the percentage error in the area is$

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$\frac{5 π}{36} \cot 63^{\circ}$

$\frac{π}{36} \cot 63^{\circ}$

$\frac{2 π}{36} \cot 63^{\circ}$

$\frac{3 π}{36} \cot 63^{\circ}$

answer is A.

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Detailed Solution

$\begin{array}{l} A_{r} = \frac{1}{2} b c \sin A, A = 63^{\circ}, Δ A = 15' \\ \Rightarrow Δ A_{r} = \frac{15}{60} \times \frac{π}{180} \\ \Rightarrow \frac{Δ A_{r}}{A_{r}} \times 100 = \cot A . Δ A \times 100 \\ = \frac{5 π}{36} \cot 63^{\circ} \end{array}$