Question

in triangle abc, \\( \angle a = 25 ^ { \circ } \\), \\( \angle c = 55 ^ { \circ } \\), and \\( ab = 60 \\). what are the approximate measures of the remaining side lengths of the triangle?
\\( \bigcirc bc \approx 31, ac \approx 72 \\)
\\( \bigcirc bc \approx 72, ac \approx 31 \\)
\\( \bigcirc bc \approx 23, ac \approx 87 \\)
\\( \bigcirc bc \approx 87, ac \approx 23 \\)

Explanation:

Step1: Find angle B

In a triangle, the sum of angles is 180°. So, ∠B = 180°−∠A−∠C.
∠B = 180°−25°−55° = 100°.

Step2: Apply the Law of Sines

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\), where \(a\), \(b\), \(c\) are the side lengths opposite to angles \(A\), \(B\), \(C\) respectively.
We know \(AB = c = 60\), \(∠A = 25°\), \(∠B = 100°\), \(∠C = 55°\).
For \(BC=a\): \(\frac{a}{\sin A}=\frac{c}{\sin C}\), so \(a=\frac{c\sin A}{\sin C}\).
\(a=\frac{60\times\sin25°}{\sin55°}\approx\frac{60\times0.4226}{0.8192}\approx31\).
For \(AC = b\): \(\frac{b}{\sin B}=\frac{c}{\sin C}\), so \(b=\frac{c\sin B}{\sin C}\).
\(b=\frac{60\times\sin100°}{\sin55°}\approx\frac{60\times0.9848}{0.8192}\approx72\).

Answer:

BC≈72, AC≈31