- #1

chwala

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- Homework Statement
- Solve the problem in the given cyclic quadrilateral

- Relevant Equations
- cyclic quadrilateral

now for part ##19.1##,

My approach is as follows, using cosine rule;

##DF= r^2 + r^2- 2r^2 cos E##

We know that angle ##E## + angle ## ∅##= ##180^0##

## ∅## is acute, therefore angle ##E## would be negative. (If ## ∅=60^0## for e.g then it follows that ##E=120^0##) Thus we shall have,

##DF^2= r^2 + r^2+2r^2 cos∅ ##

##DF^2=2r^2+2r^2 cos ∅##

##DF= \sqrt {r^2(2+2 cos ∅)}##

##DF= r\sqrt {(2+2 cos ∅)}## is the approach correct?

Now for part ##19.2##, ...this was a bit confusing to me...but i went ahead and proved lhs=rhs

To prove ##2 sin^2 ∅=1 + cos ∅##

##2 sin^2 ∅=2(1- cos^2 ∅)=2(1+cos ∅)(1-cos ∅)## =##1 + cos ∅##

##2(1-cos ∅)=1##

##1-cos ∅=\frac {1}{2}##

→##cos ∅##=##\frac {1}{2}##

on subsitituting, ##cos ∅##=##\frac {1}{2}## on both sides of the equation, we have,

##2(1-\frac {1}{4})=1+ \frac {1}{2}##

##\frac {3}{2}##=##\frac {3}{2}## thus proved. Is this correct approach or there is a better way to prove this?

For part ##19.3##

If ##cos ∅##=##\frac {1}{2}##, then ##∅= 60^0##

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