A triangle has two corners with angles of $\frac{π}{4}$ $pi / 4$ and $\frac{3 π}{8}$ $(3 pi )/ 8$ . If one side of the triangle has a length of $15$ $15$ , what is the largest possible area of the triangle?

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A triangle has two corners with angles of $\frac{π}{4}$ $pi / 4$ and $\frac{3 π}{8}$ $(3 pi )/ 8$ . If one side of the triangle has a length of $15$ $15$ , what is the largest possible area of the triangle?

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Answer 1 · 2017-10-17T05:14:34

Largest possible area of triangle $A = 135.7995$ $A = 135.7995$

Explanation:

Three angles are $\frac{π}{4}, \frac{3 π}{8}, \frac{3 π}{8}$ $pi/4, (3pi)/8, (3pi)/8$
It’s an isosceles triangle with side opposite to vertex with $(\frac{π}{4})$ $(pi/4)$ will provide the largest area possible as it is the least angle of the three.
$\frac{a}{sin a} = \frac{b}{sin b} = \frac{c}{sin c}$ $a/sin a = b/ sin b = c/ sin c$
$\frac{15}{sin (\frac{π}{4})} = \frac{b}{sin (\frac{3 π}{8})} = \frac{c}{sin (\frac{3 π}{8})}$ $15/ sin (pi/4) = b / sin ((3pi)/8) = c / sin ((3pi)/8)$
Side $b = c = \frac{15 \cdot sin (\frac{3 π}{8})}{sin (\frac{π}{4})}$ $b = c = (15*sin ((3pi)/8))/sin (pi/4)$
Side $b = c = 19.5984$ $b = c = 19.5984$
Height $h = (\frac{1}{2}) \cdot 15 \cdot tan (\frac{3 π}{8}) = 18.1066$ $h = (1/2)*15*tan ((3pi)/8) = 18.1066$
Area of triangle $A = (\frac{1}{2}) \cdot b \cdot h = (\frac{1}{2}) \cdot 15 \cdot 18.1066 = 135.7995$ $A = (1/2)*b*h = (1/2)*15*18.1066=135.7995$

Aliter :
Area of triangle $A = (\frac{1}{2}) \cdot b \cdot h$ $A = (1/2)*b*h$
As it’s an isosceles triangle, altitude bisects the base of the triangle.
$h = (\frac{1}{2}) \cdot b \cdot tan (\frac{3 π}{8})$ $h=(1/2)*b*tan ((3pi)/8)$
Area of triangle $A = (\frac{1}{2}) \cdot 15 \cdot (\frac{1}{2}) \cdot 15 \cdot tan (\frac{3 π}{8}) = 135.7995$ $A = (1/2)*15*(1/2)*15*tan((3pi)/8) = 135.7995$

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A triangle has two corners with angles of $\frac{π}{4}$ $pi / 4$ and $\frac{3 π}{8}$ $(3 pi )/ 8$ . If one side of the triangle has a length of $15$ $15$ , what is the largest possible area of the triangle?

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Explanation:

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A triangle has two corners with angles of pi / 4 π4 pi / 4 and (3 pi )/ 8 3π8 (3 pi )/ 8 . If one side of the triangle has a length of 15 1515 , what is the largest possible area of the triangle?

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A triangle has two corners with angles of $\frac{π}{4}$ $pi / 4$ and $\frac{3 π}{8}$ $(3 pi )/ 8$ . If one side of the triangle has a length of $15$ $15$ , what is the largest possible area of the triangle?