Calculate side DC in this triangle

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The discussion focuses on calculating the length of side DC in a triangle with given dimensions AD=48cm and BD=17cm. Participants confirm that the calculations appear correct and suggest using the sine and cosine formulas for verification. Additionally, they recommend checking the result by using BC, BD, and the 42° angle. Overall, the calculations are validated, and the approach is deemed appropriate.
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Homework Statement
AD=48cm
BD=17cm
Calculate DC
Relevant Equations
Sine formula and cosine formula
1587315528457.png

Was just wondering if someone could take a look at my calculations and see if I've done them correctly.
1587319123492.png
 
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VitaminK said:
Homework Statement:: AD=48cm
BD=17cm
Calculate DC
Relevant Equations:: Sine formula and cosine formula

View attachment 260979
Was just wondering if someone could take a look at my calculations and see if I've done them correctly.
View attachment 260981
That looks fine.

Of course, you could have found length DC using BC, BD, and angle C (the 42° angle). Maybe, use this as a check.
 
SammyS said:
That looks fine.

Of course, you could have found length DC using BC, BD, and angle C (the 42° angle). Maybe, use this as a check.
thanks!
 
The working out suggests first equating ## \sqrt{i} = x + iy ## and suggests that squaring and equating real and imaginary parts of both sides results in ## \sqrt{i} = \pm (1+i)/ \sqrt{2} ## Squaring both sides results in: $$ i = (x + iy)^2 $$ $$ i = x^2 + 2ixy -y^2 $$ equating real parts gives $$ x^2 - y^2 = 0 $$ $$ (x+y)(x-y) = 0 $$ $$ x = \pm y $$ equating imaginary parts gives: $$ i = 2ixy $$ $$ 2xy = 1 $$ I'm not really sure how to proceed from here.