How do you find specific angles of a triangle?

Question

Thanks megster! That's easier than I thought.

Answer 1

First you have to find the hypotenuse (use Pythagoreans theorem: a^2 + b^2 = c^2).  So take 18^2 plus 27^2 which is 1053 (324 + 729).  1053 is c^2..to find c, take the square root of 1053 (which is 32.45).

Now you have all your lengths, and to find the angles you have to use sin, cos, and tan (remember, SOH CAH TOA).  Hope this helps..

Answer 2

is to do with pythagorus theorum

he said basically

the side u want)^2 = 18^2 +27^2

the side u want = √(18^2 +27^2 )

the side u want =√324+729

= √1053

= 32.45inches to 2 decimal places

Answer 3

First find its hypotenuse and go from there..

Answer 4

First find the hypotenuse:

a^2+b^2=c^2

18^2+27^2=c^2

324+729=c^2

1053=c^2

take the square root of each side (to cancel out the squared on the c)

32.45=c

Then use trig to find the angles

sin=opp./hyp.

cos=adj./hyp.

tan=opp./adj.

So if hyp.=32.45, opp.=18 and adj.=27

What we know: Angle "C"=90 degrees

Find Angle "A"

sinA=18/32.45

sin^-1 (18/32.45)=A   (inverse functions to find angles)

A=33.69 degrees

And to fin agnle "B" you can do the same thing (but use cos) or take the easy way out and use the properties of a triagnle

you know that all 3 angles of a tigangle =180 degrees

so 180-90-33.69=B

B=56.31 degrees