Question

Solve completely #36?

Answer 1

`color(red)(hatL = 31.81^@, hatM = 36.19^@, vec(LM) = 20.1`

Explanation:

Let `x = hat M`

`hatL = 180 - 112 - x = 68-x`

`l / sin L = m / sin M`

`10.1 / Sin (68-x) = 12.8 / sin x = n / sin 112`

`sin(68-x) / sin x = 10.1/12.8 = 0.8926`

`(sin 68 cosx - cos68 sin x) / sin x = 0.8926`

`sin68cotx = 0.8926 + cos68 = 1.2672`

`cot x = 1.2672 / sin 68 = 1.3667`

`tan x = 1/1.3667 = 0.7317`

`x = tan^(-1) 0.7317 = 36.19^@`

`hatM = 36.19^@, hatL = 68 - 36.19 = 31.81^@`

Next to find `vec(LM) = n`

`n = (12.8 * sin 112) / sin x = (12.8 * sin 112) / sin 36.19 = color (green)(20.1)`

Answer 2

`LM= 19.13, hatL = 29.3° and hatM = 38.7°`

Explanation:

To solve a triangle means to find all the unknown sides and angles,

In this triangle we are given 2 sides and the angle between them (the included angle)

This means we can use the Cosine Rule to find side `ML`, called `n`

`n^2 = l^2+m^2 -2lmCosN`

`n^2 = 12.9^2+10.1^2-2(12.9)(10.1)cos112°`

`n^2 = 366.034`

`n = 19.13`

Now we can use the Sin Rule because we are working with two pairs of sides and the angles opposite them. Find the smallest angle first because there is no doubt that it will be acute.

`(sinL)/10.1 = (sin112°)/19.13`

`sinL = (10.1 sin112°)/19.13`

`sin L = 0.4895" "larr` find arcsin

`hatL = 29.3°`

Now use the sum of the angles being `180°` to find `hatM`

`hatM = 180°-112°-29.3°`

`hatM = 38.7°`