Question

On a station 'O', three horizontal angles were measure closing the horizon which are

$\angle A={34}^{\circ }{10}^{\prime }{20}^{\prime \prime }\pm 3^{\prime \prime }$

$\angle B={176}^{\circ }{40}^{\prime }{32}^{\prime \prime }\pm 4^{\prime \prime }$

$\angle C={149}^{\circ }{9}^{\prime }{4}^{\prime \prime }\pm 5^{\prime \prime }$

Then the correct value of angle A is

• A. ${34}^{\circ }{10}^{\prime }{20.52}^{\prime \prime }$
• B. ${34}^{\circ }{10}^{\prime }{20.82}^{\prime \prime }$
• C. ${34}^{\circ }{10}^{\prime }{20.62}^{\prime \prime }$
• D. ${34}^{\circ }{10}^{\prime }{20.72}^{\prime \prime }$

Answer 1

The correct option is D ${34}^{\circ }{10}^{\prime }{20.72}^{\prime \prime }$

Sum of angles A, B and C

$i.e.\angle A+\angle B+\angle C={359}^{\circ }{59}^{\prime }{56}^{\prime \prime }$

Error = M.V - TV = ${359}^{\circ }{59}^{\prime }{56}^{\prime \prime }-{360}^{\circ }{00}^{\prime }{00}^{\prime \prime }=-0^{\circ }{0}^{\prime }{4}^{\prime \prime }$

Correction = + 4''

$Letusassume{C}_1,C_2,C_{3}arecorrectionin\angle A,\angle Band\angle C$

$C_1+C_2+C_3=4"...\left(i\right)$

$C_1:C_2:C_3=3^2:4^2:5^2$

$\Rightarrow \frac{C_1}{C_2}=\frac{9}{16}\frac{C_1}{C_3}=\frac{9}{25}$

From equation (i), $C_1+\frac{16}{9}{C}_1+\frac{25}{9}{C}_1=4^{\prime \prime }\Rightarrow C_1={0.72}^{\prime \prime }$

$\therefore$ Corrected angle $\angle A={34}^{\circ }{10}^{\prime }{20.72}^{\prime \prime }$