in triangle pqr right angled at q , qr=9 and pr-pq = 1.determine the value of sinr + cosr
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in a triangle pqr where q is right angled
pq^2+qr^2=pr^2
pr^2-pq^2=qr^2
pr^2-pq^2=9^2
(pr+pq)(pr-pq)=81
(pr+pq)=81/1=81... (1)
and (pr-pq)=1...... (2)
adding 2pr=81+1=82
pr=41
and pq= 81-41=40
sinr=40/41 and cosr=9/41
sinr+cosr=49/41
pq^2+qr^2=pr^2
pr^2-pq^2=qr^2
pr^2-pq^2=9^2
(pr+pq)(pr-pq)=81
(pr+pq)=81/1=81... (1)
and (pr-pq)=1...... (2)
adding 2pr=81+1=82
pr=41
and pq= 81-41=40
sinr=40/41 and cosr=9/41
sinr+cosr=49/41
Answered by
1
Answer:
pr^2-pq^2=9^2
(pr+pq)(pr-pq)=81
(pr+pq)=81/1=81... (1)
and (pr-pq)=1...... (2)
Step-by-step explanation:
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