1. A.Use law of cosines. cosA=(b^2+c^2-a^2)/(2bc) because A is the included angle between b and c. Plug in A=50 degrees, b=13, c=6. cos50=(13^2+6^2-a^2)/(2*13*6), a^2=104.7,a=10.2 approximately, so choose A.
2. A.Use law of cosines again. cosC=(a^2+b^2-c^2)/(2ab). C=95 degrees, a=12, b=22. Plug in, cos95=(12^2+22^2-c^2)/(2*12*22), solve the equation, c^2=674, c=26 approximately. The use law of sines to solve for angle A (also works for B), a/sinA=c/sinC, 12/sinA=26/sin95, sinA=0.46, A=arcsin(0.46)=27.6. Choose A.
3. Answer is A. Area=1/2bc*sinA, since A is the included angle between b and c. Plug in b=30, c=14, A=50 degrees, area=1/2*14*30*sin50=160. 87, so the answer is A.
4. D. As long as the sum of any two sides of the triangle is bigger than the third, the triangle exists. 240+121>263, 240+263>121, 263+121>240, so it exists. To use Heron's formula, first find the semiperimeter, (240+263+121)/2=312. A=\sqrt(312*(312-240)*(312-263)*(312-121))=14499.7 approximately, so choose D.
5. 300. The included angle between the two paths is C=49.17+90=139.17 degrees. The lengths of the two paths are a=150, b=170. c is the distance we want. Use the law of cosines, cosC=(a^2+b^2-c^2)/(2ab). Plug in, c^2=89989, c=300 approximately.
From the graph, we can see that the <em>temperature increased initially</em>, then <em>started to decrease</em>. <u>All of these NOT in a constant rate, though</u>.
Roughly, the temperature was increasing from start till around 5.5 hours. Then on the temperature decreased.
So we can eliminate B, C, and D. That leaves us with correct answer as A.