Question

I really need help with this, I got 13.7 yards as the altitude by using the law of cosines and I found the area of the triangle by using the formula 1/2(side)(side)cos

Answer 1

The length of the altitude is 9.3 yards and

The area of the triangle Δ UVW is 139.3 yd².

Explanation:

Given

WU = 22 yd

WV = 30 yd

∠ UWV = 25°

To Find:

Altitude, UM = ?

area of the Δ UVW = ?

Construction:

Draw UM perpendicular to WV, that is altitude UM to WV.

Solution:

In right triangle Δ UWM if we apply Sine to angle W we get

`\sin W = \frac{\textrm{side opposite to angle W}}{Hypotenuse}\\ \sin W=(UM)/(UW) \\`

substituting the values we get

`\sin 25 = (UM)/(22)\\0.422 = (UM)/(22) \\UM = 0.422* 22\\UM = 9.284\ yd`

Therefore, the altitude from U to WV is UM = 9.3 yd.(rounded to nearest tenth)

Now for area we have formula

`\textrm{area of the triangle UVW} = (1)/(2)* Base* Altitude \\\textrm{area of the triangle UVW} = (1)/(2)* VW * UM\\=(1)/(2)* 30* 9.284\\ =139.26\ yd^(2)`

The area of the triangle Δ UVW is 139.3 yd². (rounded to nearest tenth)