Use the Law of Cosines to find angle round to the nearest tenth if necessary).

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)

it will be a cylinder. with a diameter of 8 and 4 single pieces high

step-by-step explanation: