A "trig"cky situation
By adougall, Updated
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So some trigonometry teacher named "Elaine Dria" wants a plot of land to build a house here at Sunnyside Acres...
So? What's the issue?
Here's where she wants the plot. The issue is that she's so crazy about teaching trig that she wants a triangular shaped plot! How odd!
What? That is crazy! How will we ever calculate the area of this triangular plot if we don't have the height?
We don't need the height, silly! We just need two side lengths and the angle that connects the two.
side b (3 hm)
side c (4 hm)
The equation to find the area of a triangle without needing the height is....A = .5(side b times side c) times Sine of A.....So by plugging our numbers into the equation, we get....A = .5(3 x 4)Sin30....Since 3 times 4 is 12, and half of 12 is 6, we simply have to multiply 6 by the sine of 30!
Ahhhhh this is so so so so so hard!!! What in the world is even a sine? I can't do this!
Dude, quit throwing a temper tantrum. Sine is just a trigonometric ratio. In a right triangle, it is a ratio of the opposite leg to the angle and the hypotenuse. But for here, you can just use a calculator.
So...I used my calculator and found that the sine of 30 degrees is .5, so multiplying that by 6 is 3..... I guess the area of our triangular plot is 3 hectometers squared????
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