Angle A is 27 deg and 43' b= 5.2 find the missing angle of angle B angle and C and side a and side c?

Question

find the missing angles and sides of the right triangle ABC

Answer 1

B is the right angle...so that is 90 deg
180-90-27 deg 43'=62 deg 17 min

to find side a use sinA= a/5.2
5.2*sin 27deg 43' =2.4
use pythagorean ther to find side b
5.2^2-2.4^2=c^2
c=4.6

you can check by finding side c
using sinC = c/5.2
5.2 (sin 62 deg 17 min) = 4.6

Answer 2

I will assume that angle b is the right angle. Angle c=62 17`. Side c = cos(27 43`)*5.2=4.603. Side a = sin(27 43`)*5.2=2.418.

Answer 3

first you need to convert the angle to decimal

27 degrees and 43 minutes....43/60

27.7166667 degrees.

Since we have b=5.2 we can use tan(theta)=a/b to solve for a.

tan(27.716666667)=a/5.2 this gives

5.2*tan(27.7166667)=a= 2.73199

since a^2+b^2=c^2 we can calculate c.

2.73199^2+5.2^2=c^2. or c=sqrt(2.6668613^2+5.2^2)

c=5.87399

All we have left is the last angle:

since this is a right triangle we have 90-27.1666667=62.8333333



Jon

Answer 4

A = 27d, 43m = 23.72 (approx)
The other acute angle is then 54d, 17m = 54.28 (approx)

Let's say B is the other acute angle. (Then C is the right angle.)

You can use law of sines to find a.

(sin A)/a = (sin B)/b
(sin 23.72)/a = (sin 54.28)/5.2
(sin 23.72) * 5.2 = (sin 54.28) * a
a = (sin 23.72) * 5.2 / (sin 54.28) = 2.576

Now that you have a and b, you can find c...

a^2 + b^2 = c^2
(2.576)^2 + (5.2)^2 = c^2
6.636 + 27.04 = c^2
33.676 = c^2
5.8 = c

Answer 5

Cheater do your own homework! You are cheating yourself out of an education!

Answer 6

you gotta use the law of cosines