When those conditions are given in the diagram, What is the value of sinx? ABCD is a square.

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2 Answers
Oct 8, 2017

3/5.

Explanation:

We have, from the Right- DeltaABP, AB^2+BP^2=AP^2.

:. AP^2=(2y)^2+y^2=5y^2.

Similarly, AQ^2=5y^2, and, PQ^2=2y^2.

Applying the Cosine-Rule in DeltaAPQ, we get,

cosx=(AP^2+AQ^2-PQ^2)/(2AP*AQ),

=(5y^2+5y^2-2y^2)/(2*sqrt5y*sqrt5y)=(8y^2)/(10y^2).

rArr cosx=4/5.

Since x is acute, sinx=+sqrt(1-cos^2x)=sqrt(1-16/25), giving,

sinx=3/5.

Oct 8, 2017

sinx=3/5.

Explanation:

Here is another Solution to the Problem :

Observe that DeltaABP and DeltaADQ are congruent.

:. /_PAB=/_QAD........(1).

From the Right- DeltaABP, tan/_PAB=(PB)/(AP)=y/(2y).......(2).

:.," from "(1), and, (2), tan/_PAB=1/2=tan/_QAD.

Since, /_BAP+/_PAQ+/_QAD=pi/2, we have.

/_PAQ=pi/2-(/_BAP+/_QAD),

=pi/2-2/_BAP, i.e.,

x=pi/2-2theta, where, theta=/_BAP.

:. sinx=sin(pi/2-theta)=cos2theta=(1-tan^2theta)/(1+tan^2theta).

But, tantheta=tan/_BAP=1/2.

rArr sinx={1-(1/2)^2}/{1+(1/2)^}=(3/4)/(5/4)=3/5, as Before!