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if i have sides with each having a length of 13, 14, and 15. how big will each angle be?

 May 9, 2016

Best Answer 

 #1
avatar+33661 
+5

As follows:

 

angles

.

 May 9, 2016
 #1
avatar+33661 
+5
Best Answer

As follows:

 

angles

.

Alan May 9, 2016
 #2
avatar+26400 
+5

if i have sides with each having a length of 13, 14, and 15. how big will each angle be?

 

 

Heron's formula states that the area of a triangle whose sides have lengths a, b, and c is
\(A = \sqrt{s(s-a)(s-b)(s-c)},\)
where s is the semiperimeter of the triangle; that is,
\(s=\frac{a+b+c}{2}\)

 

\(\begin{array}{rcl} s &=& \frac{a+b+c}{2} \\ &=& \frac{13+14+15}{2} \\ &=& \frac{42}{2} \\ \mathbf{s} & \mathbf{=}& \mathbf{21} \\\\ A &=& \sqrt{s\cdot (s-a)\cdot (s-b)\cdot (s-c)} \\ &=& \sqrt{21\cdot (21-13)\cdot (21-14)\cdot (21-15)} \\ &=& \sqrt{21\cdot 8 \cdot 7 \cdot 6} \\ &=& \sqrt{7056} \\ \mathbf{A} & \mathbf{=}& \mathbf{84} \end{array} \)

 

\(\begin{array}{rcl} 2\cdot A &=& b\cdot c\cdot \sin \alpha \\ \sin \alpha &=& \frac{2\cdot A}{ b\cdot c } \\ \sin \alpha &=& \frac{2\cdot 84}{ 14\cdot 15 } \\ \alpha &=& \arcsin{ (\frac{168}{ 14\cdot 15 } ) } \\ \alpha &=& \arcsin{ (0.8) } \\ \mathbf{\alpha }&\mathbf{=}& \mathbf{53.1301023542^{\circ}} \end{array} \)

 

\(\begin{array}{rcl} 2\cdot A &=& c\cdot a\cdot \sin \beta \\ \sin \beta &=& \frac{2\cdot A}{ c\cdot a } \\ \sin \beta &=& \frac{2\cdot 84}{ 15\cdot 13 } \\ \beta &=& \arcsin{ (\frac{168}{ 15\cdot 13 } ) }\\ \beta &=& \arcsin{ (0.86153846154) }\\ \mathbf{\beta} &\mathbf{=}& \mathbf{59.4897625939^{\circ}} \end{array} \)

 

\(\begin{array}{rcl} 2\cdot A &=& a\cdot b\cdot \sin \gamma \\ \sin \gamma &=& \frac{2\cdot A}{ a\cdot b} \\ \sin \gamma &=& \frac{2\cdot 84}{ 13\cdot 14 } \\ \gamma &=& \arcsin{ (\frac{168}{ 13\cdot 14 }) }\\ \gamma &=& \arcsin{ (0.92307692308) }\\ \mathbf{\gamma }&\mathbf{=}& \mathbf{67.3801350520^{\circ}} \end{array}\)

 

laugh

 May 9, 2016

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