Answer to Question #139002 in Trigonometry for riz

Question #139002
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'
1
Expert's answer
2020-10-21T17:05:17-0400

Since the angle c is not specified, we take it as 90o. 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 \u2192 \\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= \u2212\\cot \\alpha * \\cot\\beta = \u2212\\cot(76^ \u2218 41^ \u2032 ) * \\cot(49^ \u2218 23^ \u2032 )"


"\\cos C \u2248 \u22120.203 \u2192 C = \\arccos(\u22120.203)\u2248101^ \u2218 42^ \u2032"


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


"(Q9)""\\quad \\cos\\beta=\\sin\\alpha * \\cos B \u2192\\sin\\alpha ={\\cos\\beta} \/{\\cos B} = {cos(50 ^\u221810 \u2032 )}\/{\\cos(100 ^\u2218 )}"


"\\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=\u2212cos A * cosB=\u2212cos(121^ \u2218 20^ \u2032 )*cos(42 ^\u2218 01^ \u2032)"


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


"(Q4)" "\\tan A=\\tan \\alpha * \\sin B\u2192\\tan\\alpha = \\tan A\/\\sin B = \\tan(121^ \u2218 20^ \u2032 )\/\\sin(42^ \u2218 01 ^\u2032 )\n\u200b"


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


"(Q5)" "\\tan B = \\tan\\beta * \\sin A \u2192 \\tan\\alpha =\\tan B \/ \\sin A = \\tan(42 ^\u2218 01^ \u2032 )\/sin(121^ \u2218 20^\u2032 )"


"\\tan\\beta \u2248 1.055 \u2192\\beta = \\arctan(1.055)\u224846^ \u2218 31^ \u2032 50^""


4) "c=90^\\circ\\quad \\alpha=60^\\circ 35^\u2032\\quad B=122^\\circ 18^\u2032" then,

"(Q4)" "\\tan A = \\tan\\alpha * \\sin B \u2192\\tan A = \\tan(60^ \u2218 35 ^\u2032 ) * \\sin(122^ \u2218 18^ \u2032 )"


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


"(Q9)" "\\cos\\beta = \\sin\\alpha * \\cos B \u2192 \\cos\\beta=\\sin(60^ \u2218 35^ \u2032 ) * \\cos(122 ^\u2218 18^ \u2032)"


"\\cos\\beta\u2248\u22120.4655\u2192 \\beta=\\arccos(\u22120.4655)\u2248117^ \u221842 ^\u2032"


"(Q6)" "\\tan B=\u2212\\cos\\alpha* \\tan C\u2192\\tan C=\u2212 \\tan B\/\\cos\\alpha"


"\\tan C \u2248 3.221 \u2192 C = \\arctan(3.221)\u224872 ^\u2218 45^ \u2032"



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