Question

1. classify every triangle shown below and find all sides and angles given that ( ac = 9 ), ( ad = 13 ), and ( bc = 8 )( \triangle abc ) is ______ and ____( \triangle abd ) is ____ and ____( \triangle acd ) is ____ and ______

Explanation:

Step1: Classify \(\triangle ABC\)

Since \(AC = AB\) (marked with equal - side symbols), \(\triangle ABC\) is isosceles.
Using the angle - sum property of a triangle (\(A + B + C=180^{\circ}\)) and the base - angle property of an isosceles triangle (\(\angle C=\angle ABC\)). Given \(\angle A = 70^{\circ}\), then \(\angle C=\angle ABC=\frac{180 - 70}{2}=55^{\circ}\), so it is also acute.

Step2: Classify \(\triangle ABD\)

\(AB = BC = 8\), \(AD = 13\). Using the Pythagorean theorem (if it were right - angled, but \(8^{2}+8^{2}=128
eq13^{2} = 169\)). Using the angle - sum property, \(\angle ABD = 180 - 55=125^{\circ}\) (supplementary to \(\angle ABC\)). Since \(AB
eq AD
eq BD\) ( \(BD = 8\), \(AB = 8\), \(AD = 13\)), it is scalene. And since \(\angle ABD=125^{\circ}>90^{\circ}\), it is obtuse.

Step3: Classify \(\triangle ACD\)

\(AC = 9\), \(AD = 13\), \(CD=CB + BD=8 + 8 = 16\). Since \(AC
eq AD
eq CD\), it is scalene. Using the Law of Cosines \(c^{2}=a^{2}+b^{2}-2ab\cos C\) (where \(a = 9\), \(b = 13\), \(c = 16\)) to find angles. \(\cos D=\frac{AD^{2}+CD^{2}-AC^{2}}{2AD\cdot CD}=\frac{13^{2}+16^{2}-9^{2}}{2\times13\times16}=\frac{169 + 256-81}{416}=\frac{344}{416}\approx0.827\), \(\angle D\approx34^{\circ}\), \(\cos A=\frac{AC^{2}+AD^{2}-CD^{2}}{2AC\cdot AD}=\frac{9^{2}+13^{2}-16^{2}}{2\times9\times13}=\frac{81 + 169 - 256}{234}=\frac{-6}{234}\approx - 0.0256\), \(\angle A\approx91.5^{\circ}\), \(\angle C=180-(34 + 91.5)=54.5^{\circ}\). Since \(\angle A\approx91.5^{\circ}>90^{\circ}\), it is obtuse.

Answer:

\(\triangle ABC\) is isosceles and acute; \(\triangle ABD\) is scalene and obtuse; \(\triangle ACD\) is scalene and obtuse.