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#