Question

In triangle ABC, how do you solve the right triangle given 'a' is 7 m, and the angle 'A' is 22 degrees?

Answer 1

(Assuming the right angle is at C)
`/_B = 68^@`
`c= 7/sin(22^@) meters`
`b=7/tan(22^@) meters`

Explanation:

Everything that follows is under the assumption that the right angle is `/_C`. If the intent is that `/_B` is the right angle, exchange `B` and `C` in the explanation below:

Since `/_A =22^@` and `/_C = 90^@`
and the sum of interior angles of a triangle is `180^@`
`rarr /_B = 68^@`

For the `/_A`
the opposite side is `a (=7)`m
the adjacent side is `b` and
the hypotenuse is `c`

`sin(22^@) = ("opposite")/("hypotenuse")= 7/c`

`rarr c= 7/sin(22^@)`

and

`tan(22^@) = ("opposite")/("adjacent") = 7/b`

`rarr b=7/tan(22^@)`

Note:
a calculator could be used to evaluate:
`color(white)("XXX")b=7/tan(22^@) ~=17.33`
and
`color(white)("XXX")c=7/sin(22^@) ~=18.69`