in the given fig QR = 15 cm sin x = 4/5
find the measure of PQ a
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(1).
from figure
sin x= 4/5
PR/PQ=4/5
PR=4/5PQ
PQ^2=PR^2+QR^2
. =16/25 PQ^2+15^2
. =16/25 PQ^2+225
25PQ^2=16PQ^2+5625
9PQ^2=5625
PQ=75/3
PQ=25
PR=4/5 X 25
PR=20
(2). tan<PDR=1
. tan<PDR=tan 45°
. <PDR=45°
tan45°=PR/RD
1=20/RD
RD=20
PD^2=PR^2+RD^2
. =400+400
. PD^2 =800
. PD=20(2^1/2)
Since
<P+<Q+<D=180°
90°+x+45°=180°
x=180°-135°
x=45°
LHS=tan^2 x -- 1/cos^2 x
. =tan^2 45° -- 1/cos^2 45°
. =1 -- 1/
from figure
sin x= 4/5
PR/PQ=4/5
PR=4/5PQ
PQ^2=PR^2+QR^2
. =16/25 PQ^2+15^2
. =16/25 PQ^2+225
25PQ^2=16PQ^2+5625
9PQ^2=5625
PQ=75/3
PQ=25
PR=4/5 X 25
PR=20
(2). tan<PDR=1
. tan<PDR=tan 45°
. <PDR=45°
tan45°=PR/RD
1=20/RD
RD=20
PD^2=PR^2+RD^2
. =400+400
. PD^2 =800
. PD=20(2^1/2)
Since
<P+<Q+<D=180°
90°+x+45°=180°
x=180°-135°
x=45°
LHS=tan^2 x -- 1/cos^2 x
. =tan^2 45° -- 1/cos^2 45°
. =1 -- 1/
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