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Difference between revisions of "2021 Fall AMC 12B Problems/Problem 8"

(→‎Solution 1 (Area))
(→‎Solution 1 (Area))
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<math>\sin(a)=\frac{1}{2}</math>
 
<math>\sin(a)=\frac{1}{2}</math>
  
−
As the triangle is obtuse, <math>a=150^\circle</math> only. We get <math>\boxed{\textbf{(D)} \ 150}.</math>
+
As the triangle is obtuse, <math>a=150^\circ</math> only. We get <math>\boxed{\textbf{(D)} \ 150}.</math>
  
 
~Wilhelm Z
 
~Wilhelm Z
  
 
==See Also==
 
==See Also==

Revision as of 00:39, 24 November 2021

Problem

The product of the lengths of the two congruent sides of an obtuse isosceles triangle is equal to the product of the base and twice the triangle's height to the base. What is the measure, in degrees, of the vertex angle of this triangle?

$\textbf{(A)} \: 105 \qquad\textbf{(B)} \: 120 \qquad\textbf{(C)} \: 135 \qquad\textbf{(D)} \: 150 \qquad\textbf{(E)} \: 165$

Solution 1 (Area)

Let the lengths of the two congruent sides of the triangle be $x$, then the product desired is $x^2$.

Notice that the product of the base and twice the height is $4$ times the area of the triangle.

Set the vertex angle to be $a$, we derive the equation:

$x^2=4(\frac{1}{2}x^2\sin(a))$

$\sin(a)=\frac{1}{2}$

As the triangle is obtuse, $a=150^\circ$ only. We get $\boxed{\textbf{(D)} \ 150}.$

~Wilhelm Z

See Also

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