Question

this is one of my homework problems, please work through solving the triangle with me if you can! (Rounding the side lengths to the nearest tenth and the angles to the nearest degree)

Answer 1

The triangle is given with three sides b=c=8 and a=6.

To determine the angles of the triangle , use the cosine formula.

`c^2=a^2+b^2-2ab\cos C`

So to find angle C.

`8^2=6^2+8^2-2*6*8\cos C`
`64=36+64-96\cos C`
`-36=-96\cos C`
`\cos C=(36)/(96)=(6)/(16)=(3)/(8)`
`C=\cos ^(-1)(3)/(8)`
`C=67.9^(\circ)`

Hence the angle C is 67.9 degree.

To find angle b,

`b^2=a^2+c^2-2ac\cos B`

Substitute the values to find the angle cos B.

`8^2=6^2+8^2-2*6*8cosB`
`64=36+64-96\cos B`
`-36=-96\cos B`
`\cos B=(36)/(96)`
`B=\cos ^(-1)(36)/(96)`
`B=67.9^(\circ)`

Hence the angle B is 67.9 degree.

To find the angle A ,

`a^2=b^2+c^2-2bc\cos A`

Substitute the values.

`6^2=8^2+8^2-2*8*8\cos A`
`36=64+64-128\cos A`
`36-64-64=-128\cos A`
`-92=-128\cos A`
`\cos A=(92)/(128)`
`A=\cos ^(-1)(92)/(128)`
`A=\cos ^(-1)0.71875`
`A=44.04^(\circ)`

Hence the angle A is 44.04 degree.