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1 multiple choice 4 points the side lengths of a triangle are 7, 11, and 16. what kind of triangle is this? not enough information acute obtuse right 2 multiple choice 4 points given \\( \triangle a b c \\) with \\( a = 9, b = 10 \\), and \\( m \angle b = 70 \\), what type of triangle can be drawn? an obtuse triangle, only both an acute triangle and an obtuse triangle neither an acute triangle nor an obtuse triangle an acute triangle, only
Question 1
Step1: Apply the Pythagorean theorem extension
For a triangle with side lengths \(a\), \(b\), \(c\) (\(c\) is the longest side), if \(a^{2}+b^{2} \(a^{2}+b^{2}=7^{2}+11^{2}=49 + 121=170\), \(c^{2}=16^{2}=256\). By the Law of Sines \(\frac{a}{\sin A}=\frac{b}{\sin B}\), so \(\sin A=\frac{a\sin B}{b}\). Substitute \(a = 9\), \(b = 10\), \(m\angle B=70^{\circ}\), \(\sin A=\frac{9\sin70^{\circ}}{10}\approx\frac{9\times0.9397}{10}=0.8457\), then \(A\approx57.8^{\circ}\) or \(A = 180^{\circ}- 57.8^{\circ}=122.2^{\circ}\) By the Law of Sines \(\frac{\sin A}{a}=\frac{\sin B}{b}\), \(\sin A=\frac{a\sin B}{b}\). Since \(a < b\), there are two possible values for \(A\) (one acute, one obtuse). If \(A\) is acute, \(C\) is acute. If \(A\) is obtuse (\(A>90^{\circ}\)), then the triangle is obtuse. So both an acute triangle and an obtuse triangle can be drawn.Step2: Calculate \(a^{2}+b^{2}\) and \(c^{2}\)
Since \(170<256\) (\(a^{2}+b^{2}Question 2
Step1: Use the Law of Sines
Step2: Analyze the two cases for \(\angle A\)
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- C. obtuse
- B. both an acute triangle and an obtuse triangle