Question 16

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Jambmaths question: 

In triangle XYZ $\angle XYZ={{15}^{o}},\angle XZY={{45}^{o}}$and $\left| XY \right|=7cm$.Find $\left| YZ \right|$

Option A: 

$7\sqrt{2}$

Option B: 

$7cm$

Option C: 

$14\sqrt{2}cm$

Option D: 

$\frac{7\sqrt{6}}{2}cm$

Jamb Maths Solution: 

$\begin{align}  & \hat{Y}+\hat{Z}+\hat{X}={{180}^{o}} \\ & {{15}^{o}}+{{45}^{o}}+\hat{X}={{180}^{o}} \\ & \hat{X}={{180}^{o}}-{{60}^{o}}={{120}^{o}} \\ & \text{Using sine rule} \\ & \frac{x}{\sin X}=\frac{z}{\sin Z} \\ & \frac{\left| YZ \right|}{\sin {{120}^{o}}}=\frac{7}{\sin {{45}^{o}}} \\ & \left| YZ \right|=\frac{7\sin {{120}^{o}}}{\sin {{45}^{o}}}=\frac{7\times \tfrac{\sqrt{3}}{2}}{\tfrac{1}{\sqrt{2}}}=7\times \frac{\sqrt{3}}{2}\times \sqrt{2}=\frac{7\sqrt{6}}{2} \\\end{align}$

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