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********** *** given * * θ athis ***** that * is *** ************* b ** *** opposite (opp) *** * is *** *************** to *** ***** ***** θ * 550 * * *** ******* **** * we use *** ** ***** ******************* both ***** ** **** * ******* ******* * *** ***** putting *** unknown ** *** **** **** ***** * ******** * 18354m To find * ** use which ** (SOH) 1 1 b * 32sin55b = *********** ** * *** * = ****** is *** *** * ** opp ** ** **** * we use ***** ***** ** Therefore * ********** get * ** *** cos ** ***** becoming ****** * *** ******* ** ***** *********** both **** by c **** ***** dividing ******* ** ******* * 1 * ****** ** obtain * ***** ** are ***** b ** use ***** ***** ** ******* ****** * ** *** ***** is (SOH) * 1 4 *** *** * * **** the circumference ** ******** ** ********* ** ** ***** convert to *************** * r * ******************** *** ** ***** by * = **** ****** ** ** * 2 * * ***** = C = * * ********* area ** ******** in ****** ****** ** *** ********* value ** * * 0055m will ** **** in *** ****************** is given by * * πtaking ** A * * * * * * ******** m2 6 (a) ****** 12cm (b) *** *** ***** **** ** a (from ************* ******* a ***** triangle where * ** the hypotenuse *** * *** * *** ***** *** ***** ** + ** * ******* * *** * ***** + *** * 400a2 * ******* = *** *** ***** = **** = ** ******* **** * **** * *** **** * *** **** * ** ****** ****** 4 ** ** t(s) The **** under *** ***** ** a ***** of * ********* ***** ** ** Where ****** is * *** * * ** s *** * * ** sA * * = ************ ** * *** * (m/s) * ( ** 1A = 64 m Therefore the **** under *** ***** is 64 **************
