Question

in δtuv, t = 150 inches, u = 580 inches and ∠v=105°. find the length of v, to the nearest inch.

Explanation:

Step1: Apply the Law of Cosines

$$v^2 = t^2 + u^2 - 2tu\cos(V)$$

Step2: Substitute given values

$$v^2 = 150^2 + 580^2 - 2(150)(580)\cos(105^\circ)$$

Step3: Calculate each term

$$150^2=22500,\ 580^2=336400,\ 2(150)(580)=174000$$
$$\cos(105^\circ)\approx-0.2588$$

Step4: Compute the right-hand side

$$v^2 = 22500 + 336400 - 174000(-0.2588)$$
$$v^2 = 358900 + 45031.2 = 403931.2$$

Step5: Solve for v

$$v = \sqrt{403931.2}\approx627$$

Answer:

627 inches