Difference between revisions of "2011 AMC 10B Problems/Problem 7"

(Created page with '== Problem 7 == The sum of two angles of a triangle is <math>\frac{6}{5}</math> of a right angle, and one of these two angles is <math>30^{\circ}</math> larger than the other. W…')
 
 
(3 intermediate revisions by one other user not shown)
Line 1: Line 1:
== Problem 7 ==
+
== Problem==
  
 
The sum of two angles of a triangle is <math>\frac{6}{5}</math> of a right angle, and one of these two angles is <math>30^{\circ}</math> larger than the other. What is the degree measure of the largest angle in the triangle?
 
The sum of two angles of a triangle is <math>\frac{6}{5}</math> of a right angle, and one of these two angles is <math>30^{\circ}</math> larger than the other. What is the degree measure of the largest angle in the triangle?
Line 7: Line 7:
 
== Solution ==
 
== Solution ==
  
The sum of two angles in a triangle is <math>\frac{6}{5}</math> of a right angle <math>\longrightarrow \frac{6}{5} \times 9 = 108</math>
+
The sum of two angles in a triangle is <math>\frac{6}{5}</math> of a right angle <math>\longrightarrow \frac{6}{5} \times 90 = 108</math>
  
 
If <math>x</math> is the measure of the first angle, then the measure of the second angle is <math>x+30</math>.
 
If <math>x</math> is the measure of the first angle, then the measure of the second angle is <math>x+30</math>.
 
<cmath>x + x + 30 = 108 \longrightarrow 2x = 78 \longrightarrow x = 39</cmath>
 
<cmath>x + x + 30 = 108 \longrightarrow 2x = 78 \longrightarrow x = 39</cmath>
  
Now we know the measure of two angles are <math>39^{\circ}</math> and <math>69^{\circ}</math>. By the Triangle Sum Theorem, the sum of all angles in a triangle is <math>180^{\circ},</math> so the final angle is <math>72^{\circ}</math>. Therefore, the largest angle in the triangle is <math>\boxed{\textbf{(B)} 72}</math>
+
Now we know the measure of two angles are <math>39^{\circ}</math> and <math>69^{\circ}</math>. By the Triangle Sum Theorem, the sum of all angles in a triangle is <math>180^{\circ},</math> so the final angle is <math>72^{\circ}</math>. Therefore, the largest angle in the triangle is <math>\boxed{\mathrm{(B) \ } 72}</math>
 +
 
 +
== See Also==
 +
 
 +
{{AMC10 box|year=2011|ab=B|num-b=6|num-a=8}}
 +
{{MAA Notice}}

Latest revision as of 12:11, 4 July 2013

Problem

The sum of two angles of a triangle is $\frac{6}{5}$ of a right angle, and one of these two angles is $30^{\circ}$ larger than the other. What is the degree measure of the largest angle in the triangle?

$\textbf{(A)}\ 69 \qquad\textbf{(B)}\ 72 \qquad\textbf{(C)}\ 90 \qquad\textbf{(D)}\ 102 \qquad\textbf{(E)}\ 108$

Solution

The sum of two angles in a triangle is $\frac{6}{5}$ of a right angle $\longrightarrow \frac{6}{5} \times 90 = 108$

If $x$ is the measure of the first angle, then the measure of the second angle is $x+30$. \[x + x + 30 = 108 \longrightarrow 2x = 78 \longrightarrow x = 39\]

Now we know the measure of two angles are $39^{\circ}$ and $69^{\circ}$. By the Triangle Sum Theorem, the sum of all angles in a triangle is $180^{\circ},$ so the final angle is $72^{\circ}$. Therefore, the largest angle in the triangle is $\boxed{\mathrm{(B) \ } 72}$

See Also

2011 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png