QUESTION IMAGE
Question
for each of the three triangles shown, calculate the lengths of the unlabeled sides.
ac = 5√3
Step1: Identify triangle type
This is a right - triangle (\(\angle A = 90^{\circ}\)) with \(\angle B=60^{\circ}\), \(AB = 5\), \(BC = 10\). We can use trigonometric ratios or properties of 30 - 60 - 90 triangles. In a 30 - 60 - 90 triangle, the sides are in the ratio \(1:\sqrt{3}:2\) (opposite to \(30^{\circ}\), \(60^{\circ}\), \(90^{\circ}\) respectively). Here, \(\angle C=30^{\circ}\) (since the sum of angles in a triangle is \(180^{\circ}\), \(180 - 90 - 60=30^{\circ}\)), so the side opposite \(30^{\circ}\) (\(AB\)) should be half of the hypotenuse (\(BC\)). Let's verify with trigonometry. We know that \(\tan B=\frac{AC}{AB}\).
Step2: Use tangent function
We know that \(\tan60^{\circ}=\sqrt{3}\) and \(AB = 5\). From \(\tan B=\frac{AC}{AB}\), we can solve for \(AC\). Substituting the values, we get \(\tan60^{\circ}=\frac{AC}{5}\). Since \(\tan60^{\circ}=\sqrt{3}\), then \(AC = 5\times\tan60^{\circ}=5\times\sqrt{3}=5\sqrt{3}\). Also, we can check using the Pythagorean theorem. Let's see if \(AB^{2}+AC^{2}=BC^{2}\). \(AB = 5\), \(AC = 5\sqrt{3}\), \(BC = 10\). \(5^{2}+(5\sqrt{3})^{2}=25 + 75=100\) and \(10^{2}=100\), so it satisfies the Pythagorean theorem.
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\(AC = 5\sqrt{3}\)