Question 1120749
triangle ABC is a right triangle with the right angle being at angle ABC.


AB = 24 meters.


sin(ACB) = 3/5.


since sine is opposite / hypotenuse, then sin(ACB) = 3/5 = 24/AC


since 3/5 = 24/AC, solve for AC to get AC = 24 * 5 / 3 = 40.


by pythagorus, [AB]^2 + [BC]^2 = [AC]^2 which becomes 24^2 + [BC]^2 = 40^2.


solve for BC to get BC = sqrt(40^2 - 24^2) = 32.


you have AB = 24, AC = 40, BC = 32.


cos(ACB) = adjacent / hypotenuse = BC/AC = 32/40


tan(BAC) = opposite / adjacent = BC / AB = 32/24


cos(ACB)) + tan(BAC) = 32/40 + 32/24 = 4/5 + 4/3 = 12/15 + 20/15 = 32/15.


that should be your answer.


solutions are:


AB = 24
AC = 40
BC = 32
sin(ACB) + tan(BAC) = 32/15.


you could solve by finding the angles involved but that was unnecessary.


for example:


sin(ACB) = 3/5 leads to angle ACB = 36.86989765.


since sum of angles of triangle is 180, then BAC = 180 - 90 - 36.86989765 = 53.13010235.


you have:

angle ABC = 90
angle ACB = 36.86989765
angle BAC = 53.13010235.


sin(ACB)) = c/b becomes sin(36.8689765) = 24/b


solve for b to get b = 24/sin(36.8689765) = 40


cos(ACB) = a/b bcomes cos(36.8689765) = a/40


solve for a to get a = 40 * cos(36.8689765) = 32.


cos(ACB) + tan(BAC) = cos(36.86989765 + tan(53.13010235) = 2.133333333....


the denominator is assumed to be 1.


multiply numerator and denominator of that by 15 and it becomes 32/15.


that'as the same answer i got earlier.


all answser check out.


your solutions are:


AC = 40
BC = 32
cos(ACB) + tan(BAC) = 32/15 = 2.3333333333......


my diagram is shown below.


in the diagram:


c is the same as AB
b is the same as AC
a is the same as BC.


angles shown are rounded to 1 decimal point, but calculations are made from the maximum number of decimal points for each as stored in the calculator.


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