Answer to Question #139003 in Trigonometry for riz

Question

Find the remaining parts of a quadrantal triangle (c = 90°).

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 formuleas

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

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’

$$\sin A=\tan B\times \tan (90\degree-b)$$ $$\tan B=\sin A\times\tan b$$ $$B=\tan^{-1}(\sin52\degree55'\times\tan73\degree11')\approx69\degree15'2''$$ $$\sin (90\degree-a)=\cos A\times \cos(90\degree-b)$$ $$a=\cos^{-1}(\cos52\degree55'\times\sin73\degree11')\approx54\degree44'49'$$ $$\sin (90\degree-b)=\tan (-(90\degree-C))\times \tan A$$ $$\tan C=-\dfrac{\tan A}{\cos b}$$ $$C=180\degree-\tan^{-1}(\dfrac{\tan52\degree55'}{\cos73\degree11'})\approx102\degree20'5''$$