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\u22480.669\u2192 A=\\arccos(0.669)\u224841^\u2218 59^ \u203224 ^"\n \n\u200b"

$(Q5)$ "\\tan B=\\tan\\beta * \\sin A=\\tan(76^\u2218 41^ )* \\sin(41^\u2218 59^\u2032 24 ^" )"

"\\tan B \u2248 2.826\u2192 B = \\arctan(2.826)\u224870 ^\u2218 30 ^\u203240 ^"\n \n\u200b"

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

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

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 \u2248 0.7428\u2192 \\alpha = \\arcsin(0.7428)\u224847^ \u2218 58^ \u2032 12 ^""

$(Q4)$ "\\tan A= \\tan\\alpha * \\sin B = \\tan(47^ \u2218 58^ \u2032 12^") * sin(100^ \u2218 )"

"\\tan A \u2248 1.093 \u2192 A = \\arctan(1.093)\u224847^ \u2218 32^ \u2032 24^""

$(Q1)$ "\\cos C= \u2212\\cos A * \\cos B\u2192 \\cos C=\u2212\\cos(47^ \u2218 32^ \u2032 24^" ) * \\cos(100 ^\u2218 )"

"\\cos C \u2248 0.1172 \u2192 C = \\arccos(0.1172 ) \u2248 83^ \u2218 16^ \u203212 ^""

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 \u2248 0.3863 \u2192 C = arccos(0.38630)\u224867^ \u2218 16^ \u2032 50^""

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

"\\tan \\alpha\u2248\u22122.454 \u2192 \\alpha =\\arctan(\u22122.454)\u2248112^ \u221810^ \u203210^"\n \n\u200b"

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

"\\tan\\beta \u2248 1.055 \u2192\\beta = \\arctan(1.055)\u224846^ \u2218 31^ \u2032 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 \u2248 1.499 \u2192 A =\\arctan(1.499)\u224856 ^\u2218 17^ \u2032 24 ^""

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

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

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

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