Question

Help please triangles and trig

Answer 1

`\sin L = (3)/(5)`

`\tan N = (4)/(3)`

`\cos L = (4)/(5)`

`\sin N = (4)/(5)`

Explanation:

Given

The above triangle

First, we calculate the length LM using Pythagoras theorem.

`LN^2 = LM^2 + MN^2`

`10^2 = LM^2 + 6^2`

`100 = LM^2 + 36`

Collect like terms

`LM^2 = 100 - 36`

`LM^2 = 64`

Take positive square root

`LM=8`

Solving (a): Sin L

`\sin L = (Opposite)/(Hypotenuse)`

`\sin L = (MN)/(LN)`

`\sin L = (6)/(10)`

Simplify

`\sin L = (3)/(5)`

Solving (b): tan N

`\tan N = (Opposite)/(Adjacent)`

`\tan N = (LM)/(MN)`

`\tan N = (8)/(6)`

Simplify

`\tan N = (4)/(3)`

Solving (c): cos L

This calculated as:

`\cos L = (Adjacent)/(Hypotenuse)`

`\cos L = (LM)/(LN)`

`\cos L = (8)/(10)`

Simplify

`\cos L = (4)/(5)`

Solving (d): sin N

This is calculated using:

If
`a + b = 90`

Then:
`\sin a = \cos b`

So:

`\sin N = \cos L`

`\sin N = (4)/(5)`