Question 74

PRT is a tangent to a circle with centre O, at the point R on it. Diameter SQ of the circle is produced to meet the tangent at P and QRis joined. If $$\angle QRP = 28^\circ$$, then the measure of $$\angle$$SPR is:

Solution




$$\angle QRP = 28^\circ$$

$$\angle ORP = 90^\circ$$ (radius to tangent through pt of contact)

$$\angle QRO = \angle ORP - \angle QRP $$

$$\angle QRO = 90^\circ -  28^\circ = 62^\circ$$
Now,

$$\angle QRO = \angle OQR$$ 

($$ \because$$ OQ = OR(radius))

$$\angle OQR = 62^\circ$$

$$\angle RQP + \angle OQR = 180^\circ$$

$$\angle RQP = 180^\circ - 62^\circ = 118^\circ$$

In the $$\triangle$$OPR,

$$\angle SPR + \angle RQP + \angle QRP = 180^\circ$$

$$\angle SPR = 180^\circ -  118^\circ - 28^\circ = 34^\circ$$


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