Question
If $\angle\text{A}$ and $\angle\text{B}$ are acute angles such that cos A = cos B, then show that $\angle\text{A}=\angle\text{B}.$
✓
Answer
Given:
$\cos\text{A}=\cos\text{B}\ \dots(1)$
To show: $\angle\text{A}=\angle\text{B}$
$\triangle\text{ABC}$ is as shown in figure below
Now since cos A = cos B ...... from (1)
Therefore
$\frac{\text{AC}}{\text{AB}}=\frac{\text{BC}}{\text{AB}}$
Now observe that denominator of above equality is same that is AB
Hence $\frac{\text{AC}}{\text{AB}}=\frac{\text{BC}}{\text{AB}}$ only when AC = BC
Therefore AC = BC ..... (2)
We know that when two sides of a triangle are equal, then angle opposite to the sides are also equal.
Therefore from equation (2)
We can say that
Angle opposite to side AC = Angle opposite to side BC
Therefore,
$\angle\text{B}=\angle\text{A}$
Hence, $\angle\text{A}=\angle\text{B}$
$\cos\text{A}=\cos\text{B}\ \dots(1)$
To show: $\angle\text{A}=\angle\text{B}$
$\triangle\text{ABC}$ is as shown in figure below
Now since cos A = cos B ...... from (1)
Therefore
$\frac{\text{AC}}{\text{AB}}=\frac{\text{BC}}{\text{AB}}$
Now observe that denominator of above equality is same that is AB
Hence $\frac{\text{AC}}{\text{AB}}=\frac{\text{BC}}{\text{AB}}$ only when AC = BC
Therefore AC = BC ..... (2)
We know that when two sides of a triangle are equal, then angle opposite to the sides are also equal.
Therefore from equation (2)
We can say that
Angle opposite to side AC = Angle opposite to side BC
Therefore,
$\angle\text{B}=\angle\text{A}$
Hence, $\angle\text{A}=\angle\text{B}$
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