Solve each triangle. Round your answers to the nearest tenth

Mathematics

1 answer:

ivolga24 [154]3 years ago

7 0

You have to use the Law of Cosines here, since there's no other way to solve this. it's not a right triangle, so you can't use the Pythagorean Theorem. The Law of Cosines will help us find the missing side length then we will have to use the Law of Sines to find another angle. Then after that we will use the Triangle Angle-Sum theorem to finish it off. Ready? The Law of Cosines to find side b is $b^{2} = a^{2}+ c^{2} -2ac cosB$ and fill in the info we know, which is everything but the b. $b^{2} =(21) ^{2} +(29) ^{2} -2(21)(29)cos109$ . Doing all that math gives us that side b = 40.9 or 41. Now the Law of Sines to find missing angle A or C. Let's find A. $\frac{sinA}{21} = \frac{sin109}{41}$ . That gives us that angle A is 29. Now use the fact that all triangles add up to 180 to get that angle C is 42. And you're done!

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Find the roots of h(t) = (139kt)^2 − 69t + 80

Sonbull [250]

Answer:

The positive value of $k$ will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let $h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ , roots are those values of $t$ so that $h(t) = 0$ . That is:

$19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0$ (1)

Roots are determined analytically by the Quadratic Formula:

$t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}$

$t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$

The smaller root is $t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ , and the larger root is $t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ .

$h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ has one real root when $\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0$ . Then, we solve the discriminant for $k$ :