QUESTION IMAGE
Question
solve the right triangle.
diagram: right triangle with right angle at c, vertex a with angle ( 34^circ 39 ), hypotenuse ( ab = 969 ) m, side ( bc ) labeled ( a ), side ( ac ) labeled ( b )
( b = square^circ square ) (round to the nearest integer as needed.)
( a = square ) m (round to the nearest integer as needed.)
( b = square ) m (round to the nearest integer as needed.)
Step1: Find angle B
In a right triangle, the sum of angles is \(180^\circ\), and \(\angle C = 90^\circ\), \(\angle A = 34^\circ 39'\). First, convert \(34^\circ 39'\) to decimal degrees: \(39'=\frac{39}{60}=0.65^\circ\), so \(\angle A = 34.65^\circ\). Then \(\angle B = 90^\circ - 34.65^\circ = 55.35^\circ\). Convert \(0.35^\circ\) to minutes: \(0.35\times60\approx21'\), so \(\angle B = 55^\circ 21'\).
Step2: Find side a (opposite to \(\angle A\))
Using \(\sin A=\frac{a}{c}\), where \(c = 969\) m, \(\sin(34.65^\circ)\approx\sin(34^\circ 39')\). \(\sin(34.65^\circ)\approx0.568\). So \(a = c\times\sin A = 969\times0.568\approx550\) m.
Step3: Find side b (adjacent to \(\angle A\))
Using \(\cos A=\frac{b}{c}\), \(\cos(34.65^\circ)\approx0.822\). So \(b = c\times\cos A = 969\times0.822\approx797\) m.
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\(B = 55^\circ\ 21'\)
\(a = 550\) m
\(b = 797\) m