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# Triangles

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In ABC, sin A : sin B : sin C = 5:6:7 and the perimeter is 18.  Find the area of triangle ABC.

$$\phantom{\cos A}$$

May 4, 2022

#1
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A  =   (5/18)*180  =   50°

B =    (6/18) * 180 = 60°

C  =   (7/18)* 180 =  70°

By the Law of Sines

a /sin 50  = b / sin 60                 c / sin 70  = b / sin 60

a / b =  sin 50 /sqrt(3)/2             c/b = sin 70 / sqrt (3)/2

a =  b  [ 2sin 50  / sqrt 3]            c =   b [ 2sin 70 / sqrt 3 ]

a  +  b +  c  =   180

b* 2sin 50 /sqrt 3    + b  +  b [ 2sin 70 /sqrt 3 ]   =  18

b  [  2sin 50 /sqrt 3  + 1  + 2sin70/sqrt 3 ] =  18

b =  18  /  [ 2sin50/sqrt 3  + 1 + 2sin 70 / sqrt 3 ]  ≈  6.06

a = 6.06 [ 2 sin 50 / sqrt 3]  ≈ 5.36

c =  6.06 [2 sin 70 /sqrt 3 ]  ≈ 6.58

Area =  sqrt  [ 9  (9 - 6.06) (9 - 5.36) (9 - 6.58)  ]  ≈   15.267  units   May 5, 2022
#2
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The sine rule says that

$$\displaystyle \frac{a}{\sin A} = \frac{b}{\sin B}=\frac{c}{\sin C} = 2R,$$

(where R is the radius of the circumscribing circle), so

$$\displaystyle a=2R\sin A,\\b = 2R \sin B,\\ c = 2R\sin C,$$

and since

$$\displaystyle \sin A :\sin B :\sin C$$

are in the ratio 5 : 6 : 7, it follows that a, b, and c are also in the ratio 5 to 6 to 7.

Since a + b + c = 18 it then follows that a = 5, b = 6 and c = 7.

From that, s (the semi-perimeter) = 9, so the area of the triangle is

$$\displaystyle \sqrt{9(9-5)(9-6)(9-7)}=6\sqrt{6}.$$

n.b. the angles, (in degrees to 3dp) are 44.415, 57.122 and 78.463.

May 5, 2022