The following are pairs of similar triangles:
- OBN and OTS
- ONG and OSE
- OBG and OTE
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a. From [tex]\Delta OBA\sim\Delta OTE[/tex] we have
[tex]\dfrac{OB}{OT}=\dfrac{OG}{OE}\implies\dfrac{OB}4=\dfrac{8.7}{11.6}\implies OB=3[/tex]
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b. [tex]OT=OB+BT\implies4=3+BT\implies BT=1[/tex]
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c. From [tex]\Delta OBG\sim\Delta OTE[/tex] we can find the length of BG:
[tex]\dfrac{BG}{TE}=\dfrac{OB}{OT}\implies\dfrac{BG}{12}=\dfrac34\implies BG=9[/tex]
By the law of cosines, we can find the measure of angle BOG:
[tex]BG^2=OB^2+OG^2-2OB\cdot OG\cos\angle BOG[/tex]
[tex]9^2=3^2+8.7^2-2\cdot3\cdot8.7\cos\angle BOG[/tex]
[tex]\implies m\angle BOG\approx85.95^\circ[/tex]
We use the law of sines to find the measure of angle OBG:
[tex]\dfrac{\sin\angle BOG}{BG}=\dfrac{\sin\angle OBG}{OG}\implies m\angle OBG\approx74.63^\circ[/tex]
Finally,
[tex]\sin\angle OBG=\dfrac{OA}3\implies OA\approx2.89[/tex]
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d. [tex]OU=OA+AU\implies OU\approx2.89+0.9\approx3.79[/tex]
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e. [tex]OE=OG+GE\implies11.6=8.7+GE\implies GE=2.9[/tex]
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f. See part c. [tex]BG=9[/tex]
