Math  /  Geometry

Question(b) In PQR,TPQRS\triangle \mathrm{PQR}, \mathrm{TP} \| \mathrm{QRS}. (1)* Express yy in terms of θ\theta. (2) Calculate the size of θ\theta. (c) In quadrilateral ABCE,C^1=75\mathrm{ABCE}, \hat{\mathrm{C}}_{1}=75^{\circ}, B^1=20,E^2=x,F^1=y\hat{\mathrm{B}}_{1}=20^{\circ}, \hat{\mathrm{E}}_{2}=x, \hat{\mathrm{F}}_{1}=y, AF=BF\mathrm{AF}=\mathrm{BF} and FE=FC\mathrm{FE}=\mathrm{FC}. (1) Calculate the size of xx and yy. (2) Prove that ABECA B \| E C. (d) ABCD is a parallelogram with AB=AE=EC\mathrm{AB}=\mathrm{AE}=\mathrm{EC} and E^1=60\hat{\mathrm{E}}_{1}=60^{\circ}. Calculate the size of: (1) xx (2) E^2\hat{\mathrm{E}}_{2}

Studdy Solution
To find E^2 \hat{\mathrm{E}}_{2} : Since ABCD \mathrm{ABCD} is a parallelogram, E^2 \hat{\mathrm{E}}_{2} is the exterior angle to DAB \angle DAB , E^2=DAB=180E1 \hat{\mathrm{E}}_{2} = \angle DAB = 180^\circ - \angle E_1 E^2=18060 \hat{\mathrm{E}}_{2} = 180^\circ - 60^\circ E^2=120 \hat{\mathrm{E}}_{2} = 120^\circ
Final Solution:
1. y=60 y = 60^\circ
2. θ=50 \theta = 50^\circ
3. x=52.5 x = 52.5^\circ , y=80 y = 80^\circ
4. ABEC \mathrm{AB} \parallel \mathrm{EC} is not possible.
5. x=60 x = 60^\circ
6. E^2=120 \hat{\mathrm{E}}_{2} = 120^\circ

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