2 Trigonometry Problems that I cannot solve

Question

Hi, I have two trigonometry problems that I can't figure out, and I would be very grateful for help.

1.) Determine the horizontal and vertical components of the vector v with magnitude ||v|| = 4 and direction

angle θ = 110°.

2.) Let v = <2, -1> and w = <1, -2>. Decompose v into two vectors v₁ and v₂ where v₁ is parallel to w and v₂ is orthogonal to w.

Again, thank you very much, ahead of time!

Answer 1

1) I get a different answer for y, from a simpler approach: Draw an arbitrary vector in quadrant 2 with a vertical line from the tip of the vector to the x-axis. Label the vector's length 4, the vertical side y, and the horizontal side x. Then x = 4 cos 110° = -1.368 and y = 4 sin 110° = 3.759.

2) Use the cosine of the angle between two vectors being their dot product divided by the product of their lengths. The dot product x_v•x_w + y_v•y_w = 4, and each of these has length √5 from the Pythagorean Theorem. Then cos^-1 (4/5) = 36.870°.

Make a sketch with vector w along the positive x-axis and vector v at an angle above this. Draw a vertical line from the tip of v to the x-axis. The length of components parallel to w are √5 cos(-36.870) = 1.789 and √5 sin(-36.870) = -1.342, so v₁ = ,1.789, -1.342>.

Relative to the y-axis now, components perpendicular to w are √5 cos(90 - 36.870) = 1.342 and √5 sin(90 - 36.870) = 1.789, yielding v₂ = < 1.342, 1.789>.

Answer 2

I'm figuring triangles representing pressure/power.

Angle 110° ends up in quad 2. Side "c"=4.

a^2+b^2=c^2

I'll make this into 2 rt triangles.

#1. given side "c"=4, α=20° so β=70°, γ=90°.

a/sinα=b/sinβ=c/sinγ

side b=b/.939...=c (sin of rt angleγ=1) so 4*.939=3.758....vertical

" a=c^2-b^2=a^2

16-14.128....=1.871.....^.5=1.368.......horiz