Difference between revisions of "2021 Fall AMC 12B Problems/Problem 8"
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<math>\textbf{(A)} \: 105 \qquad\textbf{(B)} \: 120 \qquad\textbf{(C)} \: 135 \qquad\textbf{(D)} \: 150 \qquad\textbf{(E)} \: 165</math> | <math>\textbf{(A)} \: 105 \qquad\textbf{(B)} \: 120 \qquad\textbf{(C)} \: 135 \qquad\textbf{(D)} \: 150 \qquad\textbf{(E)} \: 165</math> | ||
| − | ==Solution== | + | ==Solution 1== |
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| + | Let the lengths of the two congruent sides of the triangle be <math>x</math>, then the product desired is <math>x^2</math>. | ||
| + | |||
| + | Notice that the product of the base and twice the height is <math>4</math> times the area of the triangle. | ||
| + | |||
| + | Set the vertex angle to be <math>a</math>, we derive the equation: | ||
| + | |||
| + | <math>4·\frac{1}{2}x^2sin(a)=x^2</math> | ||
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| + | <math>sin(a)=\frac{1}{2}</math> | ||
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| + | As the triangle is obtuse, <math>a=\frac{5\pi}{6}</math>. We get <math>\boxed{\textbf{(D)} \ 150}.</math> | ||
Still working, please don't edit now! | Still working, please don't edit now! | ||
Revision as of 22:12, 23 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?
Solution 1
Let the lengths of the two congruent sides of the triangle be 
, then the product desired is 
.
Notice that the product of the base and twice the height is 
times the area of the triangle.
Set the vertex angle to be 
, we derive the equation:
As the triangle is obtuse, 
. We get 
Still working, please don't edit now!
~Wilhelm Z
