
Concept explainers
To find: Show that sinAa=sinBb=sinCc=12r of a circumscribing a triangle.

Answer to Problem 62AYU
The answer is
sinAa=sinBb=sinCc=12r ,
Explanation of Solution
Given information:
The given radius of the circle is r
The sides and opposite angles are a, b ,c , P, Q, R of triangle ΔPQR
Concept used:
Law of Sine:
If ABC is a triangle with sides a , b and c , then
sinAa=sinBb=sinCc
The given radius of the circle is r
The given angles and sides of circumscribing triangle ΔPQR are
ΔRPQ=AΔPQR=BΔQRP=C ,
QR=a, RP=b, PQ=c ....(1)
By using the law of sine in the circumscribing triangle ΔPQR
sinAa=sinBb=sinCc ....(2)
The diameter of circle is PP'
PP'=2r ...(3)
By using theorem-Subtended angles in the same segment of a circle are equal.
∠PP'R=∠PQR=B ....(4)
From figure ∠PRP'=90∘ ....(5)
In the circumscribing triangle ΔPP'R
By using the trigonometric ratio for right triangle ΔPP'R
sinP'=oppositehypotenuse=RPPP'
According to equations ..(1), ...(2), ..(3)
sinB=b2rsinBb=12r ...(6)
Put the values of equation ....(6) in the equation ....(2)
sinAa=sinBb=sinCc=12r
Hence, the law of Tangent is a−ba+b=tan[12(A−B)]tan[12(A+B)] .
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