In triangle ABC, how do you solve the right triangle given Side AB=17 Side BC=22?

In right #\triangle ABC#, let the legs be #AB=17# & #BC=22#

Using Pythagorean theorem, in given right triangle the hypotenuse #AC# is given as

#AC=\sqrt{AB^2+BC^2}#

#=\sqrt{17^2+22^2}#

#=\sqrt773#

#=27.803#

Now, using sine formula in right triangle to find the angle #A# as follows

#\sin A=\frac{BC}{AC}#

#\sin A=\frac{22}{\sqrt773}#

#A=\sin^{-1}(\frac{22}{\sqrt773})#

#=52.306^\circ#

Similarly, using sine formula in right triangle to find the angle #C# as follows

#\sin C=\frac{AB}{AC}#

#\sin A=\frac{17}{\sqrt773}#

#A=\sin^{-1}(\frac{17}{\sqrt773})#

#=37.694^\circ#

Hence, all three sides & angles of right #\triangle ABC# are

#AB=17, BC=22, AC=\sqrt773, #

#A=52.306^\circ, B=90^\circ, C=37.694^\circ#

Let Opposite , adjacent sides of the right triangle be

#AB=c=17 and BC=a=22# and #/_B=90^0#

#AC=b# is hypotenuse , In right triangle ,

# a^2+c^2=b^2 or b^2= 17^2+22^2=773# or

#b= sqrt 773 ~~ 27.8:. AC=27.8 # unit .

The sine rule states if #a, b and c# are the lengths of the sides

and opposite angles are #A, B and C# in a triangle, then:

#a/sin A = b/sin B=c/sin C :. b/sin B=c/sin C# or

#27.8/sin 90 = 17 /sin C or sin C= 17/27.8 ; [sin 90=1]# or

#sin C~~ 0.611448 or /_C~~ sin^-1( 0.611448)# or

#/_C~~ 37.69^0 :. /_A= 180-(90+37.69)~~ 52.31^0#

Solution: # AB=17,BC=22,AC~~ 27.8# and

# /_A=52.31^0, B=90^0,/_C=37.69^0# [Ans]