Question

MEDALS MEDALS MEDALS! And Fan!
consider he right triangle ABC given below:

Part1: use the sine ratio to find the length of side b to two decimal places.

Part2: you can find the length of side a to two decimal places using the method of your choice

Answer 1

Part 1:\[\sin (25) = b/12\]\[b=12*.42\]\[b=5.04\]Part 2:\[\cos (25) = a/12\]\[a=12*.91\]\[a=10.92\]

Answer 2

okay, so im looking at it and its making sense...and so does part two. That was easier than i made it out to be! Thank you!

Answer 3

do you mind walking me through one more? @puzzler7

Answer 4

Sure. What is it?

Answer 5

Solve the triangle below by finding all missing sides and/or angles.

Part I: Use the law of cosines to find the length of side AC. Round your answer to the nearest hundredth.

Part II: Use the law of sines to find the measure of angle C.

Part III: Find the measure of angle A using any method.

Answer 6

@puzzler7 there we go

Answer 7

Part 1: (I will be using b as the side opposite B, a as side opposite A, etc.)\[b^{2}=a^{2}+c^{2}-2ac*\cos(b)\]\[b^{2}=64+100-160\cos(110)\]\[b^{2}=164+54.72\]\[b=\sqrt{218.72}\]\[b=14.79\]Part 2:\[\frac{ \sin(A) }{ a }=\frac{\sin(B)}{b}=\frac{\sin(C)}{c}\]\[\frac{\sin(C)}{10}=\frac{\sin(110)}{14.79}\]\[\sin(C)=.68\sin(110)\]\[\sin(C) = .64\]\[C=\sin^{-1} (.64)\]\[C=39.79\]Part 3: \[A+110+39.79 = 180\]\[A=180-110-34.79\]\[A=35.21\]

Answer 8

thank you for including the formulas, im making sure to write them down!