the resultant of two equal forces p making an angle theta is given by
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The resultant of two equal forces P making an angle θ, is given by. 2 P sin θ /2.
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Solution
In case I :
∣
∣
∣
∣
P
+
Q
∣
∣
∣
∣
=
P
2
+Q
2
+2Pθcosθ
⇒(2MH)
P
2
+Q
2
=
P
2
+Q
2
+2PQcosθ
⇒(2M+1)
2
(P
2
+Q
2
)=P
2
+Q
2
+2PQcosθ
⇒
2(PQ)
(4m
2
+4m)
(P
2
+Q
2
)=cosθ
In case II
∣
∣
∣
P
2
+Q
2
∣
∣
∣
=
P
2
+Q
2
+2PQsinθ
⇒(2m−1)
P
2
+Q
2
=
P
2
+Q
2
+2PQsinθ
⇒(2m−1)
2
(P
2
+Q
2
)=P
2
+Q
2
+2PQsinθ
⇒
2(PQ)
(4m
2
−4m)
(P
2
+Q
2
)=sinθ
So, tanθ=
cosθ
sinθ
=
(2PQ)
4m
2
−4m
(4m
2
+4m)
(2PQ)
(P
2
+Q
2
)
(P
2
+Q
2
)
⇒tanθ=
(4m)
(4m)
(m+1)
(m−1)
=
m+1
m−1
Here Q=α So, tanα=
m+1
m−1
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