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

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

Walleye Group TitleBest ResponseYou've already chosen the best response.0
my brain is farting
 3 years ago

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

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