Answer to Question #138993 in Trigonometry for jay nerves

Question

Find the remaining parts of a quadrantal triangle (c = 90°). Write your answers on the space provided or use a separate sheet.
1.) a = 70°10’ b = 52°40’
2.) a = 116°53’ A = 122°39’
3.) b = 69°29.7’ B = 63°4.6’
4.) a = 106°38’ b = 36°49’
5.) A = 52°55’ b = 73°11’

Answer 1

In case giving a spherical triangle in which one side is 90° , we use the following fundamental rules

$\sin middle=\tan(adj.)\times\tan(adj.)$

$\sin middle=\cos(opp.)\times\cos(opp.)$

1.) a = 70°10’ b = 52°40’ c = 90°

$\sin (-(90\degree-C))=\tan(90\degree- b)\times \tan(90\degree-a)$ $\cos C=-\cot a\times\cot b$

$C=\cos^{-1}(-\cot(70\degree10')\cot(52\degree40')\approx105\degree58'4''$

$\sin (90\degree-b)=\cos B\times \cos(90\degree-a)$ $\cos B=\dfrac{\cos b}{\sin a}$

$B=\cos^{-1}(\dfrac{\cos 52\degree40'}{\sin 70\degree10'})\approx49\degree51'27''$

$\sin (90\degree-a)=\cos A\times \cos(90\degree-b)$ $\cos A=\dfrac{\cos a}{\sin b}$

$A=\cos^{-1}(\dfrac{\cos 70\degree10'}{\sin 52\degree40'})\approx64\degree44'28''$

2.) a = 116°53’ A = 122°39’ c = 90°

$\sin (90\degree-a)=\cos A\times \cos(90\degree-b)$ $\sin b=\dfrac{\cos a}{\cos A}$

$b=\sin^{-1}(\dfrac{\cos 116\degree53'}{\cos 122\degree39'})\approx56\degree56'35''$

$\sin B=\tan A\times \tan (90\degree-a)$

$B=\sin^{-1}(\dfrac{\tan 122\degree39'}{\tan116\degree53'}\ )\approx52\degree17'50''$

$\sin (-(90\degree-C))=\tan (90\degree-b)\times \tan (90\degree-a)$ $\cos C=-\dfrac{1}{\tan a\times\tan b}$

$C=\cos^{-1}(-\dfrac{1}{\tan 116\degree53'\times\tan 122\degree39'})\approx108\degree57'21''$

3.) b = 69°29.7’ B = 63°4.6’ c = 90°

$\sin A=\tan B\times \tan (90\degree-b)$

$A=\sin^{-1}(\dfrac{\tan63\degree4.6'}{\tan69\degree29.7'})\approx47\degree25.6'$

$\sin (90\degree-b)=\cos B\times \cos(90\degree-a)$

$\sin a=\dfrac{\cos b}{\cos B}$

$a=\sin^{-1}(\dfrac{\cos69\degree29.7'}{\cos63\degree4.6'})\approx50\degree40.8'$

$\sin B=\cos (90\degree-b)\times \cos(-(90\degree-C))$ $\sin C=\dfrac{\sin B}{\sin b}$

$C=\sin^{-1}(\dfrac{\sin63\degree4.6'}{\sin69\degree29.7'})\approx72\degree9.7'$

4.) a = 106°38’ b = 36°49’ c = 90°

$\sin (-(90\degree-C))=\tan(90\degree- b)\times \tan(90\degree-a)$ $\cos C=-\cot a\times\cot b$

$C=\cos^{-1}(-\cot(106\degree38')\cot(36\degree49')\approx66\degree28'41''$

$\sin (90\degree-b)=\cos B\times \cos(90\degree-a)$ $\cos B=\dfrac{\cos b}{\sin a}$

$B=\cos^{-1}(\dfrac{\cos 36\degree49'}{\sin 106\degree38'})\approx49\degree51'27''$

$\sin (90\degree-a)=\cos A\times \cos(90\degree-b)$ $\cos A=\dfrac{\cos a}{\sin b}$

$A=\cos^{-1}(\dfrac{\cos 106\degree38'}{\sin 36\degree49'})\approx118\degree32'$

5.) A = 52°55’ b = 73°11’

$\sin A=\tan B\times \tan (90\degree-b)$ $\tan B=\sin A\times\tan b$

$B=\tan^{-1}(\sin52\degree55'\times\tan73\degree11')\approx69\degree15'2''$

$\sin (90\degree-a)=\cos A\times \cos(90\degree-b)$

$a=\cos^{-1}(\cos52\degree55'\times\sin73\degree11')\approx54\degree44'49'$

$\sin (90\degree-b)=\tan (-(90\degree-C))\times \tan A$

$\tan C=-\dfrac{\tan A}{\cos b}$

$C=180\degree-\tan^{-1}(\dfrac{\tan52\degree55'}{\cos73\degree11'})\approx102\degree20'5''$