A triangle has sides A, B, and C. If the angle between sides A and B is $\frac{π}{8}$ $(pi)/8$ , the angle between sides B and C is $\frac{π}{2}$ $(pi)/2$ , and the length of B is 3, what is the area of the triangle?

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Answer 1 · 2018-06-14T11:16:03

Area of the triangle is $1.86$ $1.86$ sq.unit.

Angle between sides $A and B$ $A and B$ is $∠ c = \frac{π}{8} = \frac{180}{8} = {22.5}^{0}$ $/_c= pi/8=180/8=22.5^0$

Angle between sides $B and C$ $B and C$ is $/_a= pi/2=180/2=90^0 :.$

Angle between sides $C and A$ is

$/_b= 180-(22.5+90)=67.5^0$ The sine rule states if

$A, B and C$ are the lengths of the sides and opposite angles

are $a, b and c$ in a triangle, then, $A/sina = B/sinb=C/sinc$

$B=3 :. A/sin a=B/sin b :. A/sin 90=3/sin 67.5$

$:. A= 3 * (sin 90/sin 67.5) ~~ 3.25 (2dp)$

Now we know sides $A~~3.25 , B=3$ and their included angle

$/_c = 22.5^0$ . Area of the triangle is $A_t=(A*B*sin c)/2$

$:.A_t=(3.25*3*sin 22.5)/2 ~~ 1.86$ sq.unit

Area of the triangle is $1.86$ sq.unit [Ans]

A triangle has sides A, B, and C. If the angle between sides A and B is (pi)/8π8(pi)/8, the angle between sides B and C is (pi)/2π2(pi)/2, and the length of B is 3, what is the area of the triangle?