Question
In a $\triangle\text{ABC},$ if $\angle\text{A}=120^\circ$ and $\text{AB}=\text{AC}.$ Find $\angle\text{A}$ and $\angle\text{C}.$
✓
Answer
Consider a $\triangle\text{ABC}.$
Given Mat $\angle\text{A}=120^\circ$ and $\text{AB}=\text{AC}$ and given
to find $\angle\text{B}$ and $\angle\text{C}.$
We can observe that $\triangle\text{ABC}$ is an isosceles triangle since $\text{AB}=\text{AC}$
$\angle\text{B}=\angle\text{C}$ (i)
[Angles opposite to equal sides are equal]
We know that sum of angles in a triangle is equal to 180°
$\Rightarrow\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{A}+\angle\text{B}+\angle\text{B}=180^\circ$
$\Rightarrow120^\circ+2\angle\text{B}=180^\circ$
$\Rightarrow2\angle\text{B}=180^\circ-120^\circ$
$\Rightarrow\angle\text{B}=\angle\text{C}=30^\circ$
Given Mat $\angle\text{A}=120^\circ$ and $\text{AB}=\text{AC}$ and given
to find $\angle\text{B}$ and $\angle\text{C}.$
We can observe that $\triangle\text{ABC}$ is an isosceles triangle since $\text{AB}=\text{AC}$
$\angle\text{B}=\angle\text{C}$ (i)
[Angles opposite to equal sides are equal]
We know that sum of angles in a triangle is equal to 180°
$\Rightarrow\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{A}+\angle\text{B}+\angle\text{B}=180^\circ$
$\Rightarrow120^\circ+2\angle\text{B}=180^\circ$
$\Rightarrow2\angle\text{B}=180^\circ-120^\circ$
$\Rightarrow\angle\text{B}=\angle\text{C}=30^\circ$
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Simplify:
$\Big(\frac{\text{x}^{\text{a}+\text{b}}}{\text{x}^\text{c}}\Big)^{\text{a}-\text{b}}\Big(\frac{\text{x}^{\text{b}+\text{c}}}{\text{x}^\text{a}}\Big)^{\text{b}-\text{c}}\Big(\frac{\text{x}^{\text{c}+\text{a}}}{\text{x}^\text{b}}\Big)^{\text{c}-\text{a}}$
→$\Big(\frac{\text{x}^{\text{a}+\text{b}}}{\text{x}^\text{c}}\Big)^{\text{a}-\text{b}}\Big(\frac{\text{x}^{\text{b}+\text{c}}}{\text{x}^\text{a}}\Big)^{\text{b}-\text{c}}\Big(\frac{\text{x}^{\text{c}+\text{a}}}{\text{x}^\text{b}}\Big)^{\text{c}-\text{a}}$
Find the area of a parallelogram given in also find the length of the altitude from vertex A on the side DC.
In the adjoining figure, ABCD is a parallelogram in which $\angle\text{A}=70^{\circ}.$ Calculate $\angle\text{B},\angle\text{C}$ and $\angle\text{D}.$
→Draw the graph of the equations given below. Also, find the coordinates of the points where the graph cuts the coordinate axes:
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