Question

In a triangle ABC , angle B= 120 Length of side b = 33 Sum of other 2 sides is 36.Find the area of triangle

Answer 1

Answer:

$area \: = \frac{27}{4} \sqrt{1219}$

Step-by-step explanation:

Calculation of side length

Le x and 36-x be the two unknown sides AB and BC respectively.

Using cosine formula, we have

$\cos( {120}^{o} ) = \frac{ {x}^{2} + {(36 - x)}^{2} - {33}^{2} }{2x(36 - x)}$

$- \frac{1}{2} = \frac{ {x}^{2} + {(36 - x)}^{2} - {33}^{2} }{2x(36 - x)}$

$- x(36 - x) = {x}^{2} + {(36 - x)}^{2} - {33}^{2}$

$- 36x + {x}^{2} = 2 {x}^{2} - 72x + {36}^{2} - {33}^{2}$

${x}^{2} - 36x + 207 = 0$

$x = \frac{36 + - \sqrt{( {36}^{2} - 4 \times 207)}}{2 \times 1}$

$x = \frac{36 + - \sqrt{468} }{2}$

$x = \frac{36 + - \sqrt{( {2}^{2} \times {3}^{2} \times 13 )} }{2}$

$x = \frac{36 + - 6 \sqrt{13} }{2}$

$x = 18 + - 3 \sqrt{13}$

Now

$when \: x = 18 + 3 \sqrt{13}$

$36 - x = 36 - (18 + 3 \sqrt{13)}$

$= 18 - 3 \sqrt{13}$

$when \: x = 18 - 3 \sqrt{13}$

$36 - x = 36 - (18 - 3 \sqrt{13})$

$= 18 + 3 \sqrt{13}$

Hence the two sides are 18+3√13 and 18 -3√13.

Calculation of area

By Heron's Formula

$area \: = \sqrt{(s)(s - a)(s - b)(s - c)}$

$s = \frac{1}{2}(33+36) = \frac{69}{2}$

$s-a = \frac{69}{2} - (18 - 3\sqrt{13})= \frac{33}{2} + 3\sqrt{13}$

$s-b = \frac{69}{2} - 33 =\frac{3}{2}$

$s-c = \frac{69}{2} - (18 + 3\sqrt{13})= \frac{33}{2} - 3\sqrt{13}$

$area \: = \sqrt{ \frac{69}{2} \frac{3}{2}( \frac{69}{2} + 3 \sqrt{13})( \frac{69}{2} - 3 \sqrt{13} ) }$

$\frac{3}{4} \sqrt{23(69 + 6 \sqrt{13})(69 - 6 \sqrt{13}) }$

$= \frac{3}{4} \sqrt{23( {69}^{2} - {6}^{2} .13)}$

$= \frac{3}{4} \sqrt{23 \times 4293}$

$= \frac{3}{4} \sqrt{23 \times {3}^{2} \times {3}^{2} \times 53 }$

$= \frac{3 \times 3 \times 3}{4} \sqrt{1219}$

$= \frac{27}{4} \sqrt{1219}$

$= 235.67$