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Question
In the ∆PQR, right angled at Q, QR=9 cm and PR-PQ =1 cm. Determine the value of sin R + cos R.
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Solution
Given values:
QR=9 cm and PR-PQ= 1 cm
Solution:
PQ=PR - 1...........[Eq1]
PR² = PQ² + QR²
PR² =(PR - 1)²+ (9)²
PR² = PR² -2PR + 1 +81
PR² - PR² +2PR = 82
PR=82/2=41cm,
PQ=40 cm
SinR=PQ/PR =40/41
CosR=QR/PR =9/41
SinR+CosR=40/41+9/41 =49/41
QR=9 cm and PR-PQ= 1 cm
Solution:
PQ=PR - 1...........[Eq1]
PR² = PQ² + QR²
PR² =(PR - 1)²+ (9)²
PR² = PR² -2PR + 1 +81
PR² - PR² +2PR = 82
PR=82/2=41cm,
PQ=40 cm
SinR=PQ/PR =40/41
CosR=QR/PR =9/41
SinR+CosR=40/41+9/41 =49/41
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