Special triangles help please?

Question

Find the value of x in these special triangles.

http://tinypic.com/view.php?pic=4uwrh37
(Click on the link, once it pops up, click on it again to make it big.)

Please help me! I can't figure these out for anything! It's driving me crazy. Thank you!

Answer 1

All these questions are based on using the trig ratios for angles of 30, 45 and 60 degrees. Use the following principles:

To find exact ratios for angles of 30 and 60 degrees, draw an equilatral triangle ABC (all angles 60deg), and make each side length 2.
Drop a perpendicular from A on to BC, meeting it at D. This then bisects angle BAC, and it also bisects the side BC.
Therefore:
DAC = 30deg.
CD = 1.

In triangle ADC, Pythagoras' Theorem proves that:
AD = sqrt(AC^2 - CD^2)
= sqrt(2^2 - 1^2)
AD = sqrt(3).
You now have all the sides and angles in this triangle, and can write down all six ratios for each of the angles 30 and 60 degrees.

For an angle of 45 degrees, draw an isosceles triangle ABC with a right-angle at B, and angles A and C each equal to 45deg.
Let the length of AB and of BC be 1.
Then AC = sqrt(2).
You can now write down all the ratios for an angle of 45 degrees.

Answer 2

All the problems use the properties of triangles and the Pythagorean Theorem. Everything is based on right triangles.

1, A right triangle with 45 degree angles has two sides equal.

Sides a and b are 4 each. x is the hypotenuse to solve for x, add the squares of the two sides and solve for square root of the sum. 4 squared is 16 therefore,

16 + 16 = 32

x = square root of 32

2. Number 4 gives you a clue to the answer of number 2.

The sides of a triangle that has 60 degrees as it's angles are all the same length. Number 2 is half of such a triangle, therefore, x = 2 times the base (5) or 10

3. This is a variation of 1. We know that the sides are the same length (x) and the hypotenuse is 14 to solve for x we need to set up the equation that allows us to substitute what we know.

x squared + x squared = 14 squared

2 x^2 = 14^2

2x^2 = 196

x^2 = 98

x = square root 98

4. A variation of 3.

6^2 + x^2 = 12^2

36 + x^2 = 144

x^2 = 144 - 36 (108)

x = square root 108

Answer 3

1. the missing side has a length of 4 since tan(45) is 1. so x = SQRT(4^2 + 4^2) = 4SQRT(2)

or you can say cos(45) = 4/x = SQRT(2)/2
x = 8/SQRT(2) = 4SQRT(2)

2. The cos(60) is 1/2 so x = 10

3. Like the first one only this time the missing side has a length of x. So:
14^2 = x^2 + x^2 = 2x^2
x^2 = 98 and x = SQRT(98) = 7SQRT(2)

or you can say:
sin(45) = x/14 = SQRT(2)/2
x = 7SQRT(2)

4. the triangle has all sides the same length ()12) so each angle is 60 degrees (180/3).
sin(60) = x/12 = SQRT(3)/2
x = 6SQRT(3)

Answer 4

specific, U can ......... truthfully those 3,4 and 5 variety a impressive angled triangle and such pairs are observed as trigonometric tuples so their ratio 3:4:5 bares comparable ratio consistently subsequently 9:12:15 = 3:4:5 it is real........

Answer 5

1) it is an isosceles triangle. therefore by pythagoras theorem x=(4^2+4^2)^0.5=32^0.5

2)cos60=5/x
x=5/cos60

3)sin45=x/14
x=14sin45

4)pythagoras theorem
x=(12^2-6^2)^0.5=108^0.5

Answer 6

1) x = 4 sqr(2)
the rest I can not see