Question

right triangle relationships and trigonometry. javier is asked to create triangles using three of four given sticks. the sticks measure 3 in., 6 in., 7 in., and 8 in. he creates these 4 triangles. triangle 1: 3 in., 6 in., 7 in. triangle 2: 3 in., 6 in., 8 in. triangle 3: 3 in., 7 in., 8 in. triangle 4: 6 in., 7 in., 8 in. how many of his triangles are obtuse?

Explanation:

Step1: Recall the Pythagorean theorem

For a right - triangle, \(a^{2}+b^{2}=c^{2}\), where \(c\) is the longest side (hypotenuse).

Step2: Check Triangle 1: 3 in., 6 in., 7 in.

\(3^{2}+6^{2}=9 + 36=45\), \(7^{2}=49\). Since \(3^{2}+6^{2}
eq7^{2}\), it is not a right - triangle.

Step3: Check Triangle 2: 3 in., 6 in., 8 in.

\(3^{2}+6^{2}=9 + 36 = 45\), \(8^{2}=64\). Since \(3^{2}+6^{2}
eq8^{2}\), it is not a right - triangle.

Step4: Check Triangle 3: 3 in., 7 in., 8 in.

\(3^{2}+7^{2}=9+49 = 58\), \(8^{2}=64\). Since \(3^{2}+7^{2}
eq8^{2}\), it is not a right - triangle.

Step5: Check Triangle 4: 6 in., 7 in., 8 in.

\(6^{2}+7^{2}=36 + 49=85\), \(8^{2}=64\). Since \(6^{2}+7^{2}
eq8^{2}\), it is not a right - triangle.

Answer:

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