Answer to Question #138991 in Trigonometry for jay nerves

Question

Solve the following quadrantal spherical triangle.

1.)a= 49° 23' , b = 76° 41'

2.) B = 100° , b = 50° 10' 

3.) A = 121° 20' , B = 42° 01 '

4.) a = 60° 35' , B = 122° 18'  

Answer 1

We will use Napier's formulas for a spherical right-angled triangle.

(More information: https://en.wikipedia.org/wiki/Spherical_trigonometry)

Hint : Since it is not specified which side is $90^\circ$ , I chose it myself.

In our case,

1) $c=90^\circ\quad a= 49^\circ 23'\quad b = 76^\circ 41'$

Then,

$$(Q8)\quad\cos a=\sin b\cdot\cos A\to\cos A=\frac{\cos a}{\sin b}=\frac{\cos(49^\circ23')}{\sin(76^\circ41')}\\[0.3cm] \cos A\approx0.669\to\boxed{ A=\arccos(0.669)\approx41^\circ 59'24''}\\[0.3cm] (Q5)\quad\tan B=\tan b\cdot\sin A=\tan(76^\circ41')\cdot\sin(41^\circ 59'24'')\\[0.3cm] \tan B\approx2.826\to\boxed{B=\arctan(2.826)\approx70^\circ30'40''}\\[0.3cm] (Q10)\quad\cos C=-\cot a\cdot\cot b=-\cot(76^\circ41')\cdot\cot(49^\circ 23')\\[0.3cm] \cos C\approx-0.203\to\boxed{C=\arccos(-0.203)\approx101^\circ42'}\\[0.3cm]$$

2) $c=90^\circ\quad B = 100^\circ\quad b = 50^\circ10'$

Then,

$$(Q9)\quad\cos b=\sin a\cdot\cos B\to\sin a=\frac{\cos b}{\cos B}=\frac{\cos(50^\circ10')}{\cos(100^\circ)}\\[0.3cm] \sin a\approx0.7428\to\boxed{a=\arcsin(0.7428)\approx47^\circ58'12''}\\[0.3cm] (Q4)\quad\tan A=\tan a\cdot\sin B=\tan(47^\circ58'12'')\cdot\sin(100^\circ)\\[0.3cm] \tan A\approx1.093\to\boxed{A=\arctan(1.093)\approx47^\circ32'24''}\\[0.3cm] (Q1)\quad\cos C=-\cos A\cdot \cos B\to\cos C=-\cos(47^\circ32'24'')\cdot\cos(100^\circ)\\[0.3cm] \cos C\approx0.1172\to\boxed{C=\arccos(0.1172)\approx83^\circ16'12''}$$

3) $c=90^\circ\quad A = 121^\circ20'\quad B = 42^\circ01'$

Then,

$$(Q1)\quad\cos C=-\cos A\cdot\cos B=-\cos(121^\circ20')\cdot\cos(42^\circ01')\\[0.3cm] \cos C\approx0.3863\to\boxed{C=\arccos(0.38630)\approx67^\circ16'50''}\\[0.3cm] (Q4)\quad\tan A=\tan a\cdot\sin B\to\tan a=\frac{\tan A}{\sin B}=\frac{\tan(121^\circ20')}{\sin(42^\circ01')}\\[0.3cm] \tan a\approx-2.454\to\boxed{a=\arctan(-2.454)\approx 112^\circ10'10''}\\[0.3cm] (Q5)\quad\tan B=\tan b\cdot\sin A\to\tan a=\frac{\tan B}{\sin A}=\frac{\tan(42^\circ01')}{\sin(121^\circ20')}\\[0.3cm] \tan b\approx1.055\to\boxed{b=\arctan(1.055)\approx 46^\circ31'50''}\\[0.3cm]$$

4) $c=90^\circ\quad a = 60^\circ35'\quad B = 122^\circ18'$

Then,

$$(Q4)\quad\tan A=\tan a\cdot\sin B\to\tan A=\tan(60^\circ35')\cdot\sin(122^\circ18')\\[0.3cm] \tan A\approx1.499\to\boxed{a=\arctan(1.499)\approx 56^\circ17'24''}\\[0.3cm] (Q9)\quad\cos b=\sin a\cdot\cos B\to\cos b=\sin(60^\circ35')\cdot\cos(122^\circ18')\\[0.3cm] \cos b\approx-0.4655\to\boxed{b=\arccos(-0.4655)\approx 117^\circ42'}\\[0.3cm] (Q6)\quad\tan B=-\cos a\cdot\tan C\to\tan C=-\frac{\tan B}{\cos a}\\[0.3cm] \tan C\approx3.221\to\boxed{C=\arctan(3.221)\approx72^\circ45'}$$

Q.E.D.