Question 44

loading...

Prev/Next links

waecmaths question: 

In the diagram,  O is the centre of the circle $\overline{PR}$is a tangent to the circle at Q and $\angle SOQ={{86}^{\circ }}$Calculate the value of $\angle SQR$

   

 

Option A: 

43o

Option B: 

47o

Option C: 

54o

Option D: 

86o

waecmaths solution: 

\[\begin{align}  & \left| PO \right|=\left| OS \right|\text{   }\!\!\{\!\!\text{ radius of a circle }\!\!\}\!\!\text{ } \\ & \angle OPS=\angle OSP\text{     }\!\!\{\!\!\text{ Base }\angle s\text{ of Iss}\text{. }\vartriangle \text{ }\!\!\}\!\!\text{ } \\ & x+x+{{86}^{\circ }}\text{=18}{{\text{0}}^{\circ }}\text{    }\!\!\{\!\!\text{ sum of }\angle s\text{ in a }\vartriangle \text{ }\!\!\}\!\!\text{ } \\ & x={{47}^{\circ }} \\ & \angle SQR={{90}^{\circ }}-x={{90}^{\circ }}-{{47}^{\circ }}={{43}^{\circ }}\text{    }\!\!\{\!\!\text{ tangent to a circle }\!\!\}\!\!\text{ } \\\end{align}\]

maths year: 
maths topics: