nycmathguy Joined Jun 27, 2021 Messages 5,386 Reaction score 422 Jan 31, 2022 #1 David Cohen Just do part (a). I will parts (b & c).
MathLover1 Joined Jun 27, 2021 Messages 2,989 Reaction score 2,884 Jan 31, 2022 #2 a. BC is hypothenuse sin(theta)=5/BC BC =5/sin(theta)
nycmathguy Joined Jun 27, 2021 Messages 5,386 Reaction score 422 Jan 31, 2022 #3 MathLover1 said: a. BC is hypothenuse sin(theta)=5/BC BC =5/sin(theta) Click to expand... Let me see. Part (b) For the little triangle above we use cosine. cos(theta) = 4/AB AB•cos(theta) = 4 AB = 4/cos(theta) Part (c) Here we use the sine function. From A to the right angle inside the little triangle, I will call that distance x. From that right angle to the one directly beneath, the distance is 5. sin(theta) = (x + 5)/(AB + BC) Note AB + BC = AC. sin(theta) = (x + 5)/AC AC•sin(theta) = (x + 5) AC = (x + 5)/sin(theta) Yes?
MathLover1 said: a. BC is hypothenuse sin(theta)=5/BC BC =5/sin(theta) Click to expand... Let me see. Part (b) For the little triangle above we use cosine. cos(theta) = 4/AB AB•cos(theta) = 4 AB = 4/cos(theta) Part (c) Here we use the sine function. From A to the right angle inside the little triangle, I will call that distance x. From that right angle to the one directly beneath, the distance is 5. sin(theta) = (x + 5)/(AB + BC) Note AB + BC = AC. sin(theta) = (x + 5)/AC AC•sin(theta) = (x + 5) AC = (x + 5)/sin(theta) Yes?