Answer to Question #139002 in Trigonometry for riz

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

Since the angle c is not specified, we take it as 90^o. So we have:

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

$(Q8)$ $\cos \alpha=\sin \beta * \cos A → \cos A = {\cos \alpha}/{\sin\beta} = {\cos 49^\circ 23'}/{\sin76^\circ 41'}$

$\cos A≈0.669→ A=\arccos(0.669)≈41^∘ 59^ ′24 ^" ​$

$(Q5)$ $\tan B=\tan\beta * \sin A=\tan(76^∘ 41^ )* \sin(41^∘ 59^′ 24 ^" )$

$\tan B ≈ 2.826→ B = \arctan(2.826)≈70 ^∘ 30 ^′40 ^" ​$

$(Q10)$ $\cos C= −\cot \alpha * \cot\beta = −\cot(76^ ∘ 41^ ′ ) * \cot(49^ ∘ 23^ ′ )$

$\cos C ≈ −0.203 → C = \arccos(−0.203)≈101^ ∘ 42^ ′$

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

$(Q9)$$\quad \cos\beta=\sin\alpha * \cos B →\sin\alpha ={\cos\beta} /{\cos B} = {cos(50 ^∘10 ′ )}/{\cos(100 ^∘ )}$

$\sin\alpha ≈ 0.7428→ \alpha = \arcsin(0.7428)≈47^ ∘ 58^ ′ 12 ^"$

$(Q4)$ $\tan A= \tan\alpha * \sin B = \tan(47^ ∘ 58^ ′ 12^") * sin(100^ ∘ )$

$\tan A ≈ 1.093 → A = \arctan(1.093)≈47^ ∘ 32^ ′ 24^"$

$(Q1)$ $\cos C= −\cos A * \cos B→ \cos C=−\cos(47^ ∘ 32^ ′ 24^" ) * \cos(100 ^∘ )$

$\cos C ≈ 0.1172 → C = \arccos(0.1172 ) ≈ 83^ ∘ 16^ ′12 ^"$

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

$(Q1)$ $cos C=−cos A * cosB=−cos(121^ ∘ 20^ ′ )*cos(42 ^∘ 01^ ′)$

$cos C ≈ 0.3863 → C = arccos(0.38630)≈67^ ∘ 16^ ′ 50^"$

$(Q4)$ $\tan A=\tan \alpha * \sin B→\tan\alpha = \tan A/\sin B = \tan(121^ ∘ 20^ ′ )/\sin(42^ ∘ 01 ^′ ) ​$

$\tan \alpha≈−2.454 → \alpha =\arctan(−2.454)≈112^ ∘10^ ′10^" ​$

$(Q5)$ $\tan B = \tan\beta * \sin A → \tan\alpha =\tan B / \sin A = \tan(42 ^∘ 01^ ′ )/sin(121^ ∘ 20^′ )$

$\tan\beta ≈ 1.055 →\beta = \arctan(1.055)≈46^ ∘ 31^ ′ 50^"$

4) $c=90^\circ\quad \alpha=60^\circ 35^′\quad B=122^\circ 18^′$ then,

$(Q4)$ $\tan A = \tan\alpha * \sin B →\tan A = \tan(60^ ∘ 35 ^′ ) * \sin(122^ ∘ 18^ ′ )$

$\tan A ≈ 1.499 → A =\arctan(1.499)≈56 ^∘ 17^ ′ 24 ^"$

$(Q9)$ $\cos\beta = \sin\alpha * \cos B → \cos\beta=\sin(60^ ∘ 35^ ′ ) * \cos(122 ^∘ 18^ ′)$

$\cos\beta≈−0.4655→ \beta=\arccos(−0.4655)≈117^ ∘42 ^′$

$(Q6)$ $\tan B=−\cos\alpha* \tan C→\tan C=− \tan B/\cos\alpha$

$\tan C ≈ 3.221 → C = \arctan(3.221)≈72 ^∘ 45^ ′$