Find all the missing parts to the triangle below
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We have:
We are given angle F as:
m∠GF=87°
Now, g=28 and h=15 we are asked to find f.
Using the law of cosines we have:
[tex]f^2=g^2+h^2-2gh\cos F\\\\f^2=(28)^2+(15)^2-2\times 28\times 15\cos (87)\\\\\\f^2=965.037796\\\\f=31.065[/tex]
which is approximately equal to 31 units
Hence, f=31 units
Also,
[tex]g^2=f^2+h^2-2fh\cos G\\\\\\28^2=15^2+31^2-2\times 15\times 31\times \cos G\\\\\\\cos G=0.4322\\\\G=\arccos 0.4322\\\\\\G=64.392[/tex]
Hence, to the nearest degree we get:
m∠G=64°
Also, we know that the sum of all the angles of a triangle is 180°
Hence,
m∠F+m∠G+m∠H=180°
i.e.
87°+64+m∠H=180°
m∠H=29°