Question

From the following figure, prove that $\theta +\varphi ={90}^{\circ }$. Also prove that there are two other right angled triangle. Find $\sin \alpha ,\cos \beta and\tan \varphi .$

Answer 1

To prove that $\theta +\varphi ={90}^{\circ }$, we check if triangle ABC is right angle or not using pythagoras theorem.
AB = 9+16 = 25 (hypotenuse)
AC = 15 (base)
BC = 20 (perpendicular)
Now, let's check $\sqrt{(15{)}^2+(20{)}^2}=\sqrt{225+400}=\sqrt{625}=25=AB$
Thus, we can say that $\theta +\varphi ={90}^{\circ }$
In triangle ACD
Check AC =$\sqrt{A{D}^2+C{D}^2}=\sqrt{(9{)}^2+(12{)}^2}=20$
Thus, triangle BCD is right angled triangle
Now, the value of $\sin \alpha ,\cos \beta and\tan \varphi$
To find $\sin \alpha$,
In triangle ACD,
$\sin \alpha =\frac{oppositeside}{hypotenuse}=\frac{12}{15}Tofind\cos \beta IntriangleBCD,\cos \beta =\frac{adjacentside}{hypotenuse}=\frac{16}{20}=\frac{4}{5}Tofind\tan \varphi IntriangleBCD,\tan \varphi =\frac{oppositeside}{adjacentside}=\frac{16}{12}=\frac{4}{3}$