A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
DEF m<F=43, d=16mm, and f=24mm, find m<D
anonymous
 one year ago
DEF m<F=43, d=16mm, and f=24mm, find m<D

This Question is Closed

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.0\(\bf \textit{Law of sines} \\ \quad \\ \cfrac{sin(\measuredangle A)}{a}=\cfrac{sin(\measuredangle B)}{b}=\cfrac{sin(\measuredangle C)}{c}\qquad thus \\ \quad \\ \cfrac{sin(F)}{f}=\cfrac{sin(D)}{d}\implies d\cdot \cfrac{sin(F)}{f}=sin(D) \\ \quad \\ sin^{1}\left( d\cdot \cfrac{sin(F)}{f} \right)=sin^{1}[sin(D)]\implies sin^{1}\left( d\cdot \cfrac{sin(F)}{f} \right)=\measuredangle D\)

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.0hmm that's jus the law of sines notation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so what do i plug in for A

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.0\(\bf \textit{Law of sines} \\ \quad \\ \cfrac{sin(\measuredangle A)}{a}=\cfrac{sin(\measuredangle B)}{b}=\cfrac{sin(\measuredangle C)}{c}\qquad thus \\ \quad \\ \\ \cfrac{sin(F)}{f}=\cfrac{sin(D)}{d}\implies d\cdot \cfrac{sin(F)}{f}=sin(D) \\ \quad \\ sin^{1}\left( d\cdot \cfrac{sin(F)}{f} \right)=sin^{1}[sin(D)]\implies sin^{1}\left( d\cdot \cfrac{sin(F)}{f} \right)=\measuredangle D \\ \quad \\ \begin{cases} F=43^o\\ f=24\\ d=16 \end{cases}\qquad sin^{1}\left( 16\cdot \cfrac{sin(43^o)}{24} \right)=\measuredangle D\)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.