A community for students.
Here's the question you clicked on:
 0 viewing
itspriusse
 3 years ago
Find x.
A)4
B)6.7
C)7.2
D)9
itspriusse
 3 years ago
Find x. A)4 B)6.7 C)7.2 D)9

This Question is Closed

itspriusse
 3 years ago
Best ResponseYou've already chosen the best response.0@Walleye can you help me ?

Walleye
 3 years ago
Best ResponseYou've already chosen the best response.0im trying but I cant seem to think of how to get the answer

itspriusse
 3 years ago
Best ResponseYou've already chosen the best response.0@taewony can you help me ?

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1323558134983:dw ABCD is a cyclic quadrilateral, so:\[ \begin{align} \angle BCD+\angle BAD&=180^0\\ \text{but }\angle BAP+\angle BAD&=180^0\\ \therefore \angle BAP&=\angle BCD&=\alpha\\ \text{similarly, we can show:}\\ \angle ABP &= \angle ADC&=\beta\\ \text{now, using the sine rule we get:}\\ \text{in }\triangle PAB\frac{3}{\sin(\alpha)}&=\frac{4}{\sin(\beta)}\tag{1}\\ \text{in }\triangle PDC\frac{5+4}{\sin(\alpha)}&=\frac{3+x}{\sin(\beta)}\tag{2}\\ \text{dividing (1) by (2) we get:}\\ \frac{3}{5+4}=\frac{4}{3+x}\\ \therefore 9+3x=36\\ \therefore 3x=27\\ \therefore x=9 \end{align}\]
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.