Circle inscribed

There is a triangle ABC and a circle inscribed in this triangle with a radius of 15. Point T is the point of contact of the inscribed circle with the side BC. What is the area of the triangle ABC if | BT | = 25 a | TC | = 26?

Final Answer:

S =  1170

Step-by-step explanation:

$$\begin{gathered} r=15 \\ BT=25 \\ TC=26 \\ BC=BT+TC=25+26=51 \\ BS=\sqrt{r^2+B{T}^2}=\sqrt{15^2+25^2}=5\sqrt{34}\doteq 29.1548 \\ BZ=\sqrt{B{S}^2-r^2}=\sqrt{29.1548^2-15^2}=25 \\ CS=\sqrt{r^2+T{C}^2}=\sqrt{15^2+26^2}=\sqrt{901}\doteq 30.0167 \\ CY=\sqrt{C{S}^2-r^2}=\sqrt{30.0167^2-15^2}=26 \\ A{S}^2=r^2+A{Z}^2 \\ A{S}^2=r^2+A{X}^2 \\ AX=AZ \\ \sin \beta /2=r:BS \\ \beta =\frac{180°}{\pi }\cdot 2\cdot \arcsin (r/BS)=\frac{180°}{\pi }\cdot 2\cdot \arcsin (15/29.1548)\doteq 61.9275{}^{\circ } \\ \sin \gamma /2=r:CS \\ \gamma =\frac{180°}{\pi }\cdot 2\cdot \arcsin (r/CS)=\frac{180°}{\pi }\cdot 2\cdot \arcsin (15/30.0167)\doteq 59.9633{}^{\circ } \\ \alpha =180-\beta -\gamma =180-61.9275-59.9633\doteq 58.1092{}^{\circ } \\ \tan \alpha /2=r:AY \\ AY=r/\tan (\alpha /2)=15/\tan (58.1092/2)=27 \\ AZ=AY=27 \\ AC=AY+CY=27+26=53 \\ AB=AZ+BZ=27+25=52 \\ s=(AC+AB+BC)/2=(53+52+51)/2=78 \\ S=\sqrt{s\cdot (s-AC)\cdot (s-AB)\cdot (s-BC)}=\sqrt{78\cdot (78-53)\cdot (78-52)\cdot (78-51)}=1170 \end{gathered}$$

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