Question

Help !!! Please !!!!!!!!!!!

Answer 1

We are given
`\gamma=16.7^(\circ)`, a right triangle and hypothenuse
`a=49`

Using angle function
`\cos(x)` we are able to find the value of
`t`.

From picture (and definitions) we know that

`\cos(\gamma)=t/a`

And now just solve for
`t`

`t=a\cos(\gamma)=49\cos(16.7^(\circ))\approx46.93^(\circ)`

To find
`c` use sine

`\sin(\gamma)=c/a\Longrightarrow c=a\sin(\gamma) \\ 49\sin(16.7^(\circ))\approx14.08^(\circ)`

Find
`T` by subtracting
`C` and
`90^(\circ)` from
`180^(\circ)` hence
`T=73.3^(\circ)`

The answer is A, C and D.

Hope this helps.