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Solve the Problem A=12absinC\mathrm { A } = \frac { 1 } { 2 } \mathrm { ab } \sin \mathrm { C } \text {, }

question 206

Multiple Choice

Solve the problem.
-The area of a triangle is given by
A=12absinC\mathrm { A } = \frac { 1 } { 2 } \mathrm { ab } \sin \mathrm { C } \text {, }
where a\mathrm { a } and b\mathrm { b } are the lengths of two of the sides and C\mathrm { C } is the included angle. If A=15\mathrm { A } = 15 in. 2 , a=5\mathrm { a } = 5 in., b=8\mathrm { b } = 8 in., and C\mathrm { C } is an acute angle, what must C\mathrm { C } be? Give your answer in degrees to the nearest hundredth.


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