Answer:
Step-by-step explanation:
3a)
find side HP , let's set side HP = c in the law of cosines formula
then
c = sq rt [ 23^2 + 12^2 - 2*12*23*cos(117) ]
c = sq rt [ 529 + 144 - (-250.6027559) ]
c = sq rt [ 673 + 250.6027559 ]
c = sq rt [ 923.602755 ]
c = 30.39
then
HP = 30.4 ( rounded to nearest 10th )
3b)
let PR = c, then law of cosines is (we need angle Q also 180=24+97+Q is 59°
c = sq rt [ 9^2 + 22^2 - 2*9*22*cos(59) ]
c = sq rt [ 81 + 484 -396*0.5150380 ]
c = sq rt [ 565 - 203.955 ]
c = sq rt [ 391.04495]
c = 19.7748
PR = 19.8 ( rounded to nearest 10th )
3c)
they want us to find ∠B, in the formula for cosines ∠B is really ∠C, it's a bit confusing.. I'm sure they did that on purpose to mess people up. Kind of underhanded :/ Tell the teacher that you're "appalled" at the deviousness of the test :/ stage a protest against math that teaches treachery :P storm the principals/ dean's office with paper airplanes and pictures of anime :DDD okay back to math.
a=28
b=17
c=15
since we want angle B then the formula looks like
B= arcCos ( a^2 + c^2 - b^2 / 2*a*c) so we'll avoid messing up the which side is which ... by setting up the formula this way .. got it?
B = arcCos [28^2 + 15^2 - 17^2 / 2*28*15]
B = arcCos[784 + 225 - 289 /840 ]
B = arcCos[720/ 840]
B = arcCos[0.85714285]
B = 31.0027 °
that it came out so close to an integer I'm pretty sure it's right
∠B = 31 .0 ° ( rounded to the nearest 10th )
3d)
find ∠A
set it up carefully again, they are asking us to move the variable around again
a = 18
b= 28
c= 12
then the formula looks like
∠A = arcCos[ b^2 + c^23 - a^2 / 2*b*c ]
∠A = arcCos[28^2 + 12^2 - 18^2 /2*28*12 ]
∠A = arcCos[784 + 144 -324 / 672 ]
∠A = arcCos[604/672]
∠A = arcCos[0.8988095]
∠A =25.99797 °
∠A =26.0 ° ( rounded to nearest 10th)
