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anyone know how to do #14 and #15?????

Anyone Know How To Do 14 And 15 class=

Sagot :

Answer:

He Melanie ,  use SOH  CAH  TOA to remember the trig functions

Step-by-step explanation:

for 14 they hyp = 22 and we know one angle of 45°,  we also know that the triangle has a right angle or 90°  and by inspection we can tell that the other angle is also going to be 45°   so we know X and Y will be the same length, is why this is a good observation.  

Use one of the trig functions with H in it b/c you know Hyp is 22,  so,  

SOH  =  Sin(Θ)= Opp / Hyp  

Sin(45) = Opp / 22

Sin(45) * 22 = Opp

[tex]\sqrt{2}[/tex] / 2  * 22 = opp

11*[tex]\sqrt{2}[/tex] = opp     is the exact answer,   if  you would like some decimal places, it's 7.7781 units,  both X and Y

for 15  we can use SOH again for the bigger triangle.  we know the Hyp is 24 and the angle is 30°

Sin(30) = Opp / 24

Sin(30) *24 = Opp

12 = Opp         ( Because I know Sin(30)  = 1/2 )

Opp is the upright between the two triangles

Use CAH   Cos(Θ) = Adj / Hyp

Cos(30) = Adj / 24

Cos(30) * 24 = Adj

[tex]\sqrt{3}[/tex] /2 *24 =Adj

12[tex]\sqrt{3}[/tex] = Adj

Z = 12[tex]\sqrt{3}[/tex]   or if you want some decimals 20.7846

Becasue the smaller triangle is also a 45° right triangle we know that the upright and side Y are the same lengths

then

Y = 12

You could use SOH or CAH  to find the Hyp but, lets just resort to Pythagoras,  and use  his formula to find Hyp

Hyp = [tex]\sqrt{12^{2} +12^{2} }[/tex]

Hyp = [tex]\sqrt{288}[/tex]

hyp = 16.97

X = 16.97

:)