Question

Can someone help please
ASAP

Answer 1

see explanation

Explanation:

using the tangent ratio in the right triangle

tan A =
`(opposite)/(adjacent)` =
`(BC)/(AB)` =
`(5)/(11)` , then

∠ A =
`tan^(-1)` (
`(5)/(11)` ) ≈ 24.4° ( to 1 decimal place )

tan C =
`(opposite)/(adjacent)` =
`(AB)/(BC)` =
`(11)/(5)` , then

∠ C =
`tan^(-1)` (
`(11)/(5)` ) ≈ 65.6° ( to 1 decimal place )

using Pythagoras' identity in the right triangle

the square on the hypotenuse is equal to the sum of the squares on the other two sides, that is

AC² = AB² + BC² = 11² + 5² = 121 + 25 = 146 ( take square root of both sides )

AC =
`√(146)` ≈ 12.08 ( to 2 decimal places )