Solve triangle ( A B C ) if ( angle A=43.1^circ, a=188.7 ),...

Solve triangle ( A B C ) if ( angle A=43.1^{circ}, a=188.7 ), and ( b=243.4 ). [ sin B= ] (round answer to 5 decimal places) There are two possible angles ( B ) between ( 0^{circ} ) and ( 180^{circ} ) with this value for sine. Find the two angles, and report them so that ( angle B_{1} ) is the acute angle. [ angle B_{1}= ] [ ext { and } angle B_{2}= ] (round these and all remaining answers to 1 decimal place) Thus, two triangles satisfy the given conditions: triangle ( A_{1} B_{1} C_{1} ) and triangle ( A_{2} B_{2} C_{2} ). Solve the first triangle: ( A_{1} B_{1} C_{1} ) [ angle C_{1}= ] . and ( c_{1}= ) Solve the second triangle: ( A_{2} B_{2} C_{2} ) [ angle C_{2}= ] and ( c_{2}= )

Transcript

00:01 Solve triangle abc if angle a is 43 .1 degrees, side a is 188 .7, and b is 243 .4.

00:10 So, first we want to find the sine of b.

00:14 So, we can show a, we can show the law of sines here.

00:19 The sine of b over side b, and since we're given information about side a, or angle a, that would be sine of angle a over a.

00:29 So, we can solve this for sine of b by multiplying both sides by b, and that gives us the sine of b equals b times the sine of a over a.

00:43 So, then the sine of b would be the b, which is 243 .4, times the sine of angle a, which is 43 .1, and that's over side a, which is 188 .7, okay? and we can plug that in the calculator, and keeping that in degree mode, to five decimal places, that gives us the sine of b to be about 0 .88134.

01:28 So, we'll write that right here, the sine of b to five decimal places is 0 .88134.

01:37 So, that's the first part we're gonna need.

01:42 And next, this is walking us through how to solve this, and we're gonna find the two angles between zero and 180 degrees that have this value for sine.

01:51 So, we're gonna have an acute angle, we'll call that angle b1, and an obtuse angle, we'll call that angle b2.

01:56 So, in the calculator, we know that sine of b, that the sine of b equals 0 .88134.

02:09 So, if we can solve for angle b by finding the inverse sine, of 0 .88134, and this would actually give us the acute angle, so this would be angle b1.

02:26 So, if we plug that in the calculator, the inverse sine of 0 .88134 to one decimal place, that gives us angle b1 to be 61 .8 degrees.

02:40 So, that's b1, 61 .8 degrees.

02:45 That's the first angle.

02:46 The obtuse angle will just be the supplement of that acute angle.

02:51 So, b1, angle b1 plus angle b2 have to be supplementary, they're gonna add to 180.

03:00 So, angle b2 would just be 180 minus angle b1, which is 61 .8.

03:09 So, 180 minus 61 .8, that gives us 118 .2.

03:16 So, the obtuse angle, or angle b2, would be 118 .2 degrees.

03:21 Both of those have a sine of 0 .88134.

03:28 So, this means that two triangles can be formed with the given measurements, one with angle b1 and the other with angle b2.

03:34 And the original given measurements, angle a, side a, and side b would all be the same.

03:41 So, we just need to solve for c now.

03:44 In the case where we have angle b1, angle c1 would just use, to find it, we would use the fact that all three angles have to add 180.

03:58 We know, for solving triangle a1, b1, c1, angle a1 would still be originally what angle a was, which was 43 .1.

04:10 Angle b1, we're using the acute version right now, that was 61 .8.

04:16 So, angle c1 would just be 180, since all three angles add to 180...

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