In triangle PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P
Answers
Answer:
⇒PQ=5cm
⇒PR+QR=25cm
⇒PR=25−QR
Now, In △PQR
⇒(PR)
2
=PQ
2
+QR
2
⇒(25−QR)
2
=5
2
+QR
2
⇒625+QR
2
−50QR=25+QR
2
⇒50QR=600
⇒QR=12cm
⇒PR=25−12=13cm
∴sinP=
PR
QR
=
13
12
,cosP=
PR
PQ
=
13
5
,tanP=
PQ
QR
=
5
12
Hence, the answers are sinP=
13
12
,cosP=
13
5
,tanP=
5
12
.
Answer:
Solution: Given,
In triangle PQR,
PQ = 5 cm
PR + QR = 25 cm
Let us say, QR = x
Then, PR = 25 – QR = 25 – x
Using Pythagoras theorem:
PR2 = PQ2 + QR2
Now, substituting the value of PR, PQ and QR, we get;
(25 – x)2 = (5)2 + (x)2
252 + x2 – 50x = 25 + x2
625 – 50x = 25
50x = 600
x = 12
So, QR = 12 cm
PR = 25 – QR = 25 – 12 = 13 cm
Therefore,
sin P = QR/PR = 12/13
cos P = PQ/PR = 5/13
tan P = QR/PQ = 12/5
Step-by-step explanation:
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