How would go about finding the lengths of the longer leg and hypotenuse of a 30-60-90 triangle if the shorter leg was 10 ft?
The anagram SOHCAHTOA is extremely useful in this situation. The shorter side would be across from the smallest angle. Therefore you could use the Sin(x)=opposite/hypotenuse. Sin(30)=10/h and solve for h. To find the adjacent leg you can use either TOA or CAH since you have the hypotenuse now. tan(x)=opposite/adjacent angle. tan(30)=10/a and solve for a.
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