You are given two sides of a triangle, a = 4.5 and b = 6, the angle between them is 35 degrees, how do you find the third side and the area of the triangle.?

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Answer 1 · 2018-08-01T11:27:45

Third side is $3.47$ unit and area of triangle is $7.74$ sq.unit.

Sides $a=4.5 , b=6$ and their included angle $/_C = 35^0$

Cosine rule: (for all triangles) $c^2=a^2 + b^2 − 2 a b cos(C)$

$:. c^2=4.5^2 + 6^2 − 2*4.5*6*cos(35)$ or

$c^2~~12.02 :. c ~~ 3.47 (2dp)$

Sides $a=4.5 , b=6$ and their included angle

$/_C = 35^0$ . Area of the triangle is $A_t=(a*b*sin C)/2$

$:.A_t=(4.5*6*sin 35)/2 ~~ 7.74$ sq.unit

Area of the triangle is $7.74$ sq.unit.

Third side is $3.47$ unit and area of triangle is $7.74$ sq.unit

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