How do you solve the right angle triangle given A=70°-49'-15.6", a=23, b=8?

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Answer 1 · 2017-01-14T15:39:48

#angleB=19°10'44.4'', c = 24.352#

An easy way to remember the trigonometric ratios in right angled
triangles:

Let opposite side be “t”
Let hypotenuse side be “s”
Let adjacent side be “a”

#Sin theta =(opposite)/ ( hypotenuse) = t/s = Sin(ts)#

#Cos theta = (adjacent ) /( hypotenuse ) = a/s = Cos(as)#

#Tan theta = (opposite)/(adjacent) = t/a = Tan(ta)#

#Cosec theta = (hypotenuse)/(adjacent) = s/t = Cosec(st)#

#Sec theta = (hypotenuse)/(adjacent) = s/a = Sec(sa)#

#Cot theta = (adjacent)/(opposite) = a/t = Cot(at)#

#180° -(70° 49'15.6''+90°)=19°10'44.4''= angleB#

#s/a = Sec 19°10'44.4''#

Multiply both sides by a

#s = Sec 19°10'44.4'' xx a#

#s= 1.058764789 xx 23#

#s = 24.352# = hypotenuse = side b#

Check:

#t/s = Sin 19°10'44.4'' #

Multiply both sides by s

#t= Sin 19°10'44.4'' xx s#

#t = 0.328520492 xx 24.352 #

#t = 8# = opposite side = b#