Question

in \\( \triangle a b c, \overline { b m } \\) is a median, \\( \overline { b h } \\) is an altitude, \\( b h = 6 \\), \\( m \angle b a c = 30 ^ { \circ } \\), and \\( b m = \frac { 1 } { 2 } a c \\). find \\( a b, b c \\), and \\( a c \\). answer: \\( a b = \\) \\( b c = \\) \\( a c = \\)

Explanation:

Step1: Find \(AB\) in right - triangle \(ABH\)

In right - triangle \(ABH\), \(\sin\angle BAC=\frac{BH}{AB}\). Given \(\angle BAC = 30^{\circ}\) and \(BH = 6\).
Since \(\sin30^{\circ}=\frac{1}{2}\), we have \(\frac{1}{2}=\frac{6}{AB}\).
Cross - multiply: \(AB\times1=6\times2\).

Step2: Determine \(AC\) using the property of the median

Since \(BM\) is a median (\(AM = MC\)) and \(BM=\frac{1}{2}AC\), then \(AM = BM=MC\).
In right - triangle \(ABH\), \(AB = 12\) (from Step1). Also, \(\cos\angle BAC=\frac{AH}{AB}\). Since \(\cos30^{\circ}=\frac{\sqrt{3}}{2}\), \(AH = AB\cos30^{\circ}=12\times\frac{\sqrt{3}}{2}=6\sqrt{3}\).
Because \(BM = AM\), \(\angle A=\angle ABM = 30^{\circ}\), then \(\angle BMC=\angle A+\angle ABM = 60^{\circ}\). And since \(BM = MC\), \(\triangle BMC\) is equilateral. So \(MC=BM = AM\).
Let \(AM=x\), then \(AC = 2x\). In right - triangle \(ABH\), \(AH=\sqrt{AB^{2}-BH^{2}}=\sqrt{144 - 36}=\sqrt{108}=6\sqrt{3}\).
Since \(BM=\frac{1}{2}AC\) and \(BM = AM\), in right - triangle \(ABH\), \(AM = BM\). Using the property of \(30 - 60-90\) triangle (in \(\triangle ABH\)), and the fact that \(BM = AM\), we know \(AC = 12\).

Step3: Find \(BC\)

Since \(\triangle BMC\) is equilateral (because \(BM = MC\) and \(\angle BMC = 60^{\circ}\)), \(BC=BM\). And since \(BM=\frac{1}{2}AC\) and \(AC = 12\), \(BC = 6\).

Answer:

\(AB = 12\), \(BC=6\), \(AC = 12\)