i need help about this problem,,...

hey

30-60-90 Triangles The guy who named 30-60-90 triangles didn’t have much of an imagination. These triangles have angles of , , and . What’s so special about that? This: The side lengths of 30-60-90 triangles always follow a specific pattern. Suppose the short leg, opposite the 30° angle, has length x. Then the hypotenuse has length 2x, and the long leg, opposite the 60° angle, has length x. The sides of every 30-60-90 triangle will follow this ratio of 1: : 2 . This constant ratio means that if you know the length of just one side in the triangle, you’ll immediately be able to calculate the lengths of all the sides. If, for example, you know that the side opposite the 30º angle is 2 meters long, then by using the 1: : 2 ratio, you can work out that the hypotenuse is 4 meters long, and the leg opposite the 60º angle is 2 meters. And there’s another amazing thing about 30-60-90 triangles. Two of these triangles joined at the side opposite the 60º angle will form an equilateral triangle. Here’s why you need to pay attention to this extra-special feature of 30-60-90 triangles. If you know the side length of an equilateral triangle, you can figure out the triangle’s height: Divide the side length by two and multiply it by . Similarly, if you drop a “perpendicular bisector” (this is the term the SAT uses) from any vertex of an equilateral triangle to the base on the far side, you’ll have cut that triangle into two 30-60-90 triangles. Knowing how equilateral and 30-60-90 triangles relate is incredibly helpful on triangle, polygon, and even solids questions on the SAT. Quite often, you’ll be able to break down these large shapes into a number of special triangles, and then you can use the side ratios to figure out whatever you need to know. 45-45-90 Triangles A 45-45-90 triangle is a triangle with two angles of 45° and one right angle. It’s sometimes called an isosceles right triangle, since it’s both isosceles and right. Like the 30-60-90 triangle, the lengths of the sides of a 45-45-90 triangle also follow a specific pattern. If the legs are of length x (the legs will always be equal), then the hypotenuse has length x: Know this 1: 1: ratio for 45-45-90 triangles. It will save you time and may even save your butt. Also, just as two 30-60-90 triangles form an equilateral triangles, two 45-45-90 triangles form a square. We explain the colossal importance of this fact when we cover polygons a little later in this chapter.

I hope that gives you an idea on how to solve. If not, I will be back in a little while to help you further if you would like.

hix hix... i still dont understand:((....

For trig functions, think of them as ratios or fractions. Use the acronym SOA CAH TOA, SOH => Sin(angle) = Opposite/Hypotenuse.........CAH => Cos(angle) = Adjacent/Hypotenuse..........TOA => Tan(angle) = Opposite/Adjacent...................For any right triangle, if you pick an angle, that angle is formed by the hypotenuse and the adjacent side, the 3rd side is always the opposite side...................Also need to know that sec = 1/cos, csc = 1/sin, cot = 1/tan.

yes i got that...

Lets look at 1st one....find sec(30) - cos(60) \[\sec(30) = \frac{1}{\cos(30)}\] Break it into 2 parts, find cos(30) first......cos so use CAH \[\cos(30) = \frac{adjacent}{hypotenuse}\] find the 30 degree angle on triangle, adjacent side is sqrt3, hypotenuse is 2 \[\cos(30) = \frac{\sqrt{3}}{2}\] Next find cos(60)....find 60 degree angle, adjacent is 1, hypotenuse is 2 \[\cos(60) = \frac{1}{2}\] put it all together \[\rightarrow \frac{1}{\frac{\sqrt{3}}{2}} + \frac{1}{2} = \frac{2}{\sqrt{3}}+\frac{1}{2} = \frac{2\sqrt{3}}{3}+\frac{1}{2} = \frac{4\sqrt{3}+3}{6}\]

oops should be "- 1/2" at last line

thanks for helping

i will try to do with another one....

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