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How do you find perimeter of a triangle with 1 side given the angles are 105, 45, and 30 and the only side given is 8, opposite of the angle 30?

1 Answer

Jun 12, 2016

Perimeter: $34.76852$ $34.76852$ (to the limit of my calculator)

Explanation:

Let $a$ $a$ be the length of the side opposite the $45^{\circ}$ $45^@$ angle
and $b$ $b$ the length of the side opposite the $105^{\circ}$ $105^@$ angle.

The Law of Sines tells us
$XXX \frac{8}{sin (30^{\circ})} = \frac{a}{sin (45^{\circ})} = \frac{b}{sin (105^{\circ})}$ $color(white)("XXX")8/sin(30^@)=a/sin(45^@)=b/sin(105^@)$

$sin (30^{\circ}) = 0.5$ $sin(30^@)=0.5$
$sin (45^{\circ}) = 0.707107$ $sin(45^@)=0.707107$ (approx.)
$sin (105^{\circ}) = 0.965926$ $sin(105^@)=0.965926$ (approx.)

$\frac{8}{sin (30^{\circ})} = 16$ $8/sin(30^@)=16$

$a = 16 \times sin (45^{\circ}) = 11.31371$ $a=16xxsin(45^@)=11.31371$

$b = 16 \times sin (105^{\circ}) = 15.45481$ $b=16xxsin(105^@)=15.45481$

$Perimeter = 8 + a + b$ $"Perimeter" = 8 + a + b$

$XXXXX = 34.76852$ $color(white)("XXXXX")=34.76852$

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