Welcome to Westonci.ca, where you can find answers to all your questions from a community of experienced professionals. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
Sagot :
[tex]Look\ at\ the\ picture.\\\\tan30^o=\frac{\frac{a}{2}}{\frac{a\sqrt3}{2}}=\frac{a}{2}:\frac{a\sqrt3}{2}=\frac{a}{2}\cdot\frac{2}{a\sqrt3}=\frac{1}{\sqrt3}\cdot\frac{\sqrt3}{\sqrt3}=\frac{\sqrt3}{3}[/tex]
Here is a mantra that I learned in Geometry class in high school. We were
all very excited, as angles had been invented only a few years earlier:
In a 30-60 right triangle ...
the side opposite the 30 is half the hypotenuse
the side opposite the 60 is half the hypotenuse times the square root of three.
So the tangent of the 30 would be (opposite) / (adjacent) = (1/2h) / (1/2h√3) .
That's (1)/(√3) or (√3)/(3) .
all very excited, as angles had been invented only a few years earlier:
In a 30-60 right triangle ...
the side opposite the 30 is half the hypotenuse
the side opposite the 60 is half the hypotenuse times the square root of three.
So the tangent of the 30 would be (opposite) / (adjacent) = (1/2h) / (1/2h√3) .
That's (1)/(√3) or (√3)/(3) .
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.