In a triangle PQR. Q=8cm , R=6cm . Cosp=1/12.calculatethe value of p
I think you mean:
In a triangle PQR. q=8cm , r=6cm . Cos(P)=1/12. Calculate the value of p.
where capital letters mean vertices, lower-case letters mean lengths of sides opposite the corresponding vertex.
Use cosine rule:
p²=q²+r²-2qr(cos(P))
and solve for p.
To find the value of side P in triangle PQR, we can use the Law of Cosines, which states that in a triangle with sides a, b, and c, and angle C opposite side c, the cosine of angle C is related to the sides by the formula:
c² = a² + b² - 2abcos(C)
In this case, we have:
QR = 8 cm
PR = 6 cm
cos(P) = 1/12
Let's substitute these values into the Law of Cosines formula and solve for side PQ (which equals side PR):
PR² = PQ² + QR² - 2PQ * QR * cos(P)
(6 cm)² = PQ² + (8 cm)² - 2PQ * (8 cm) * (1/12)
36 = PQ² + 64 - 4/3 * PQ
Multiplying through by 3 to eliminate the fraction:
108 = 3PQ² + 192 - 4PQ
Rearranging the terms:
3PQ² - 4PQ + 84 = 0
Now, we have a quadratic equation in terms of PQ. We can use the quadratic formula to solve for PQ:
PQ = (-b ± √(b² - 4ac)) / 2a
In our case:
a = 3
b = -4
c = 84
PQ = (-(-4) ± √((-4)² - 4 * 3 * 84)) / (2 * 3)
PQ = (4 ± √(16 - 1008)) / 6
Since the discriminant (b² - 4ac) is negative, the quadratic does not have any real solutions. Therefore, there is no real value for side P in this triangle.