Question

Solve the right triangle. B = \boxed{}^\circ \boxed{}' (Round to the nearest integer as needed.) a = \boxed{} m (Round to the nearest integer as needed.) b = \boxed{} m (Round to the nearest integer as needed.)

          Solve the right triangle.
B = \boxed{}^\circ \boxed{}' (Round to the nearest integer as needed.)
a = \boxed{} m (Round to the nearest integer as needed.)
b = \boxed{} m (Round to the nearest integer as needed.)
        
Solve the right triangle.
B = ^∘' (Round to the nearest integer as needed.)
a =  m (Round to the nearest integer as needed.)
b =  m (Round to the nearest integer as needed.)

Added by Francisco C.

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Precalculus with Limits
Precalculus with Limits
Ron Larson 2nd Edition
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Solve the right triangle. B=deg , ' (Round to the nearest integer as needed.) a=-m (Round to the nearest integer as needed.) b=-m (Round to the nearest integer as needed.) Solve the right triangle B 969 m 42025 b B= (Round to the nearest integer as needed.) m (Round to the nearest integer as needed.) m (Round to the nearest integer as needed.) a= b=
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Solve the triangle ABC, if the triangle exists. B = 137.3°, c = 7.476, b = 15.426 Select the correct choice below and fill in the answer boxes with the choice. A. There are 2 possible solutions for the triangle. The measurements for the solution with the longer side a are as follows. m∠A = (Round to the nearest tenth as needed.) m∠C = (Round to the nearest tenth as needed.) The length of side a = (Round to the nearest thousandth as needed.) The measurements for the solution with the shorter side a are as follows. m∠A = (Round to the nearest tenth as needed.) m∠C = (Round to the nearest tenth as needed.) The length of side a = (Round to the nearest thousandth as needed.) B. There is only 1 possible solution for the triangle. The measurements for the remaining angles A and C and side a are as follows. m∠A = (Round to the nearest tenth as needed.) m∠C = (Round to the nearest tenth as needed.) The length of side a = (Round to the nearest thousandth as needed.) C. There are no possible solutions for the triangle.

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Transcript

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00:01 Okay, so i'm going to just draw a triangle, not to scale, just a place to kind of put my information.
00:09 I'm going to label it abc.
00:13 Now, angle b measures 137 .3 degrees.
00:20 Side c is 7 .476, and side b is 15 .426.
00:28 And i'm going to proceed using law of signs to see what situation i have because i have side -side angle.
00:37 That's that ambiguous case and we don't know until we get started whether we have one, two, or no triangle.
00:47 So 137 .3 degrees over its corresponding side equals sign of c over 7 .476.
00:58 We're going to cross multiply and then divide both sides by 15 .426.
01:15 Okay, so let's do that.
01:17 Grab a calculator, make sure you're in degree mode, 137 .3 times 7 .476, divide that by 15 .426.
01:28 You're going to get sign of c equals 0 .3287.
01:39 So c is the inverse sign.
01:42 So let's find the inverse sign of 0 .3287.
01:49 And that's going to give us 19 .2 degrees.
01:54 Or remember, the ambiguous case, you always have to consider the supplement to that or 160 .8.
02:02 Okay, but we already have 137 in our triangle...
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