Answer to Question #126027 in Trigonometry for Habeeb goat

Question #126027

In a triangle, suppose A=37∘, a=3, and b=4.


Find B, C, and c.


SHOW ALL WORK TO RECEIVE FULL CREDIT


If there are no solutions, show why and then write NO SOLUTIONS as your final answer.


If there are two solutions, find B1, C1, and c1


and then find B2, C2, and c2


.


1
Expert's answer
2020-07-15T19:07:32-0400

1) According to the Law of cosines:

"a^2=c^2+b^2-2bc\\sdot cosA"

"3^2=c^2+4^2-2\\sdot4c\\sdot cos37^o"

"c^2-8c\\sdot0.8+16-9=0"

"c^2-6.4c+7=0"

We got a quadratic equation "ax^2+bx+c=0"

Its common solution will be 

"x_1,_2={-b\u00b1\\sqrt{b^2-4ab}\\over2a}"

In our case its particular solutions are

"c_1={{6.4+\\sqrt{6.4^2-4\\sdot1\\sdot7}}\\over{2\\sdot1}}={{{6.4+\\sqrt{12.96}}\\over{2}}={{6.4+3.6}\\over{2}}}=3.2+1.8=5"

"c_2={{6.4-\\sqrt{6.4^2-4\\sdot1\\sdot7}}\\over{2\\sdot1}}={{6.4-\\sqrt{12.96}}\\over{2}}={{6.4-3.6}\\over{2}}=3.2-1.8=1.4"

2) According to the Law of sines:

"{sinA\\over{a}}={sinB\\over{b}}={sinC\\over{c}}"

Now we will find B, C

"B_1={arcsin\u2061({{bsinA}\\over{a}})}={arcsin\u2061({{4\\sdot{3\\over{5}}}\\over{3}})}={arcsin\u2061({4\\over{5}})}=53^o"

"C_1={arcsin\u2061({{csinA}\\over{a}})}={arcsin\u2061({{5\\sdot{3\\over{5}}}\\over{3}})}={arcsin\u2061(1)}=90^o"

"B_2={arcsin\u2061({{bsinA}\\over{a}})}={arcsin\u2061({{4\\sdot{3\\over{5}}}\\over{3}})}={arcsin\u2061({4\\over{5}})}=127^o"

"C_2={arcsin\u2061({{bsinA}\\over{a}})}={arcsin\u2061({{1,4\\sdot{3\\over{5}}}\\over{3}})}={arcsin\u2061({7\\over{25}})}=16^o"


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