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Question Number 107555 by anonymous last updated on 11/Aug/20

Answered by bshahid010 last updated on 11/Aug/20

$$ifcos\alpha =-\frac{\sqrt{3}}{2}\Rightarrow sin\alpha =\frac{1}{2}$$$$andsin\beta =\frac{\sqrt{2}}{2}\Rightarrow cos\beta =\frac{\sqrt{2}}{2}$$$$Nowcos(\alpha +\beta )=cos\alpha .cos\beta -sin\alpha .sin\beta$$$$\Rightarrow -\frac{\sqrt{3}}{2}.\frac{\sqrt{2}}{2}-\frac{1}{2}.\frac{\sqrt{2}}{2}$$$$\Rightarrow -\left(\frac{\sqrt{6}-\sqrt{2}}{4}\right)$$$$$$$$$$

Commented by anonymous last updated on 12/Aug/20

thank a lot sir

Answered by bshahid010 last updated on 11/Aug/20

$$Q.2andrestcanbedoneinsimilarway$$$$\mathrm{\angle }A={20}^{°},a=12andc=31$$$$usinglawofsines$$$$\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}$$$$\Rightarrow \frac{12}{sin20°}=\frac{c}{sinC}$$$$\Rightarrow sinC=\frac{31\times sin{20}^{°}}{12}\approx 0.8835$$$$\Rightarrow C={\sin }^{-1}(0.8835)=62.06°$$$$Now\mathrm{\angle }B=180°-(\mathrm{\angle }A+\mathrm{\angle }B)$$$$\Rightarrow \mathrm{\angle }B=180°-(20+62.06)$$$$\Rightarrow \mathrm{\angle }B=97.94$$

Commented by anonymous last updated on 12/Aug/20

Thank you Sir

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