In triangle pqr angle b is equal to angle q by 2 equal to angle by 6 then find the measure of angle p
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Let PQ=xPQ=x, QR=yQR=y and PR=zPR=z.
Given: x+y+z=60x+y+z=60 (i);
Equate the areas: 12∗xy=12∗QS∗z12∗xy=12∗QS∗z (area of PQR can be calculated by 1/2*leg*leg and 1/2* perpendicular to hypotenuse*hypotenuse) --> xy=12zxy=12z (ii);
Aslo x2+y2=z2x2+y2=z2 (iii);
So, we have:
(i) x+y+z=60x+y+z=60;
(ii) xy=12zxy=12z;
(iii) x2+y2=z2x2+y2=z2.
From (iii) (x+y)2−2xy=z2(x+y)2−2xy=z2 --> as from (i) x+y=60−zx+y=60−z and from (ii) xy=12zxy=12z then (60−z)2−2∗12z=z260−z)2−2∗12z=z2 --> 3600−120z+z2−24z=z23600−120z+z2−24z=z2 --> 3600=144z3600=144z --> z=25z=25;
From (i) x+y=35x+y=35 and from (ii) xy=300xy=300 --> solving for xx and yy --> x=20x=20 and y=15y=15 (as given that x>yx>y).
Given: x+y+z=60x+y+z=60 (i);
Equate the areas: 12∗xy=12∗QS∗z12∗xy=12∗QS∗z (area of PQR can be calculated by 1/2*leg*leg and 1/2* perpendicular to hypotenuse*hypotenuse) --> xy=12zxy=12z (ii);
Aslo x2+y2=z2x2+y2=z2 (iii);
So, we have:
(i) x+y+z=60x+y+z=60;
(ii) xy=12zxy=12z;
(iii) x2+y2=z2x2+y2=z2.
From (iii) (x+y)2−2xy=z2(x+y)2−2xy=z2 --> as from (i) x+y=60−zx+y=60−z and from (ii) xy=12zxy=12z then (60−z)2−2∗12z=z260−z)2−2∗12z=z2 --> 3600−120z+z2−24z=z23600−120z+z2−24z=z2 --> 3600=144z3600=144z --> z=25z=25;
From (i) x+y=35x+y=35 and from (ii) xy=300xy=300 --> solving for xx and yy --> x=20x=20 and y=15y=15 (as given that x>yx>y).
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