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what formula would i use to solve the problem for area of this triangle: triangle ABC, a=3, angleB=24, angleC=24

Guest May 10, 2015

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 #3
avatar+1068 
+13

 triangle ABC, a=3, angleB=24, angleC=24

 

1/   You need to find the height(h) of this triangle:                                                                                                        

        $$h = tan(B)(a/2)$$

 

 2/  Now you can calculate the area of the triangle using this formula:

 

        $$Area(A) = h(a/2)$$

civonamzuk  May 11, 2015
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4+0 Answers

 #1
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Pythegorem Thereom

Guest May 10, 2015
 #2
avatar+18712 
+13

what formula would i use to solve the problem for area of this triangle: triangle ABC, a=3, angleB=24, angleC=24

$$\\(1) \qquad Area=\dfrac{a\cdot b}{2}\cdot\sin{(C)}\\\\
(2) \qquad b=a \cdot \dfrac{\sin{(C)}} {\sin{(A)}} }\\\\
(3) \qquad \sin{(A)}= \sin{( 180\ensurement{^{\circ}}-(B+C) )} = \sin{( B+C ) }$$

 

$$Area=\dfrac{a^2}{2} \cdot \dfrac{ \sin{(B)} \cdot\sin{(C)} } {\sin{(B+C)}} = \dfrac{a^2}{2} \cdot
\left( \dfrac{ 1 } { \cot{(B)} + \cot{(C)} } } \right)\\\\\\
Area = \dfrac{3^2}{2} \cdot
\left( \dfrac{ 1 } { \cot{(24\ensurement{^{\circ}})} + \cot{(24\ensurement{^{\circ}})} } } \right)\\\\
Area = \dfrac{3^2}{2} \cdot
\left( \dfrac{ 1 } { 2\cdot \cot{(24\ensurement{^{\circ}})} } } \right)\\\\
\boxed{
Area = \dfrac{3^2}{4} \cdot \tan{(24\ensurement{^{\circ}})} } } \\\\
Area = \dfrac{3^2}{4} \cdot 0.4452286853\\\\
Area = \dfrac{9}{4} \cdot 0.4452286853\\\\
Area = 1.0017645419$$

heureka  May 10, 2015
 #3
avatar+1068 
+13
Best Answer

 triangle ABC, a=3, angleB=24, angleC=24

 

1/   You need to find the height(h) of this triangle:                                                                                                        

        $$h = tan(B)(a/2)$$

 

 2/  Now you can calculate the area of the triangle using this formula:

 

        $$Area(A) = h(a/2)$$

civonamzuk  May 11, 2015
 #4
avatar+26322 
+8

civonamzuk has spotted that this is an isosceles triangle, which simplifies the calculation significantly!

.

Alan  May 11, 2015

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