How do you solve the right triangle ABC given a=3, A=26?

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Answer 1 · 2018-07-16T11:31:27

In right #\triangle ABC#, we have #\angle A=26^\circ# & #a=3# then other acute angle #\angle C# is given as

#\angle C=180^\circ-\angle B-\angle A#

#=180^\circ-90^\circ-26^\circ#

#=64^\circ#

Now, by applying Sine rule in right #\triangle ABC# as follows

#\frac{a}{\sin \angleA}=\frac{b}{\sin \angleB}=\frac{c}{\sin \angleC}#

#\frac{3}{\sin 26^\circ}=\frac{b}{\sin 64^\circ}=\frac{c}{\sin 90^\circ}#

Consider,

#\frac{3}{\sin 26^\circ}=\frac{b}{\sin 64^\circ}#

#b=\frac{3\sin64^\circ}{\sin26^\circ}#

#=6.151#

Consider,

#\frac{3}{\sin 26^\circ}=\frac{c}{\sin 90^\circ}#

#c=\frac{3\sin90^\circ}{\sin26^\circ}#

#=6.843#