Question
In a $\triangle\text{ABC}$ right angled at B, $\angle\text{A}=\angle\text{C.}$ Find the values of.
$\sin\text{A}\sin \text{B}+\cos\text{A}\cos\text{B}$
$\sin\text{A}\sin \text{B}+\cos\text{A}\cos\text{B}$
✓
Answer
Given: $\angle\text{A}=\angle\text{C}$ and $\angle\text{B}=90^\circ$ 
To find: $\sin\text{A}\sin\text{B}+\cos\text {A}\cos\text{B}\dots(1)$ Now we know that sum of all the angles of any triangle is 180°Therefore,
$\angle\text{A}+ \angle\text{B}+\angle\text{C}=180^\circ $
since $\angle\text{A}= \angle\text{C}$ and $\angle\text{B}=90^ \circ$Therefore,
$\angle\text{A}+90^ \circ+\angle\text{A}=180^\circ$ $\Rightarrow2\angle\text{A}+90^ \circ=180^\circ$ $\Rightarrow2\angle\text{A}=180^\circ- 90^\circ$ $\Rightarrow2\angle\text{A}=90^\circ$ $\Rightarrow\angle\text{A}=\frac{90^ \circ}{2}$ $\Rightarrow\angle\text{A}=45^\circ$ It is given that $\angle\text{A}=\angle\text{C}$ Therefore, $\angle\text{A}=\angle\text{C}=45^ \circ\dots(2)$ Now we have, $\sin\text{A}=\frac{\text{BC}}{\text {AC}},\ \sin\text{B}=\sin90^\circ=1$ $\cos\text{A}=\frac{\text{AB}}{\text {AC}},\ \cos\text{B}=\cos90^\circ=0$ Now by substituting the above values in equation (1) We get, $\sin\text{A}\sin\text{B}+\cos\text {A}\cos\text{B}=\sin45^\circ\times1+ \cos45^\circ\times0$ $= \sin45^\circ$ Since $\sin45^\circ=\frac {1}{\sqrt{2}}$Therefore $\sin\text{A}\sin \text{B}+\cos\text{A}\cos\text{B}=\frac {1}{\sqrt{2}}$
To find: $\sin\text{A}\sin\text{B}+\cos\text {A}\cos\text{B}\dots(1)$ Now we know that sum of all the angles of any triangle is 180°Therefore,$\angle\text{A}+ \angle\text{B}+\angle\text{C}=180^\circ $
since $\angle\text{A}= \angle\text{C}$ and $\angle\text{B}=90^ \circ$Therefore,
$\angle\text{A}+90^ \circ+\angle\text{A}=180^\circ$ $\Rightarrow2\angle\text{A}+90^ \circ=180^\circ$ $\Rightarrow2\angle\text{A}=180^\circ- 90^\circ$ $\Rightarrow2\angle\text{A}=90^\circ$ $\Rightarrow\angle\text{A}=\frac{90^ \circ}{2}$ $\Rightarrow\angle\text{A}=45^\circ$ It is given that $\angle\text{A}=\angle\text{C}$ Therefore, $\angle\text{A}=\angle\text{C}=45^ \circ\dots(2)$ Now we have, $\sin\text{A}=\frac{\text{BC}}{\text {AC}},\ \sin\text{B}=\sin90^\circ=1$ $\cos\text{A}=\frac{\text{AB}}{\text {AC}},\ \cos\text{B}=\cos90^\circ=0$ Now by substituting the above values in equation (1) We get, $\sin\text{A}\sin\text{B}+\cos\text {A}\cos\text{B}=\sin45^\circ\times1+ \cos45^\circ\times0$ $= \sin45^\circ$ Since $\sin45^\circ=\frac {1}{\sqrt{2}}$Therefore $\sin\text{A}\sin \text{B}+\cos\text{A}\cos\text{B}=\frac {1}{\sqrt{2}}$
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