Question

suppose a triangle has sides a, b, and c, and that $a^2 + b^2 < c^2$. let $\theta$ be the measure of the angle opposite the side of length c. which of the following must be true? check all that apply.

a. $\cos\theta > 0$

b. $\cos\theta < 0$

c. the triangle is not a right triangle.

d. $a^2 + b^2 - c^2 = 2ab\cos\theta$

Explanation:

Step1: Use Law of Cosines

Recall the Law of Cosines: $c^2 = a^2 + b^2 - 2ab\cos\theta$

Step2: Rearrange the formula

Rearrange to isolate $\cos\theta$:
$\cos\theta = \frac{a^2 + b^2 - c^2}{2ab}$

Step3: Analyze $\cos\theta$ sign

Given $a^2 + b^2 < c^2$, so $a^2 + b^2 - c^2 < 0$. Since side lengths $a,b>0$, $2ab>0$. Thus $\frac{\text{negative}}{\text{positive}} < 0$, so $\cos\theta < 0$.

Step4: Check right triangle condition

For a right triangle, the Pythagorean theorem $a^2 + b^2 = c^2$ holds. Here $a^2 + b^2 < c^2$, so it cannot be a right triangle.

Step5: Verify option D

Rearranging the original Law of Cosines $c^2 = a^2 + b^2 - 2ab\cos\theta$ gives $a^2 + b^2 - c^2 = 2ab\cos\theta$, which matches option D.

Answer:

B. $\cos\theta < 0$, C. The triangle is not a right triangle., D. $a^2 + b^2 - c^2 = 2ab\cos\theta$