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Question Number 180615 by cherokeesay last updated on 14/Nov/22

Answered by Acem last updated on 14/Nov/22

Commented by cherokeesay last updated on 14/Nov/22

$$thankyousir.$$

Answered by mr W last updated on 14/Nov/22

Commented by cherokeesay last updated on 14/Nov/22

$$Nice,thankyousir.$$

Commented by mr W last updated on 14/Nov/22

$$anotherway:$$$$x^2={15}^2+{\left(\sqrt{2}\sqrt{137}\right)}^2-2\times 15\times \sqrt{2}\sqrt{137}\cos (\frac{\pi }{4}+\alpha )$$$$x^2={15}^2+2\times 137-30\sqrt{137}\times (\frac{11}{\sqrt{137}}-\frac{4}{\sqrt{137}})$$$$x^2={15}^2+2\times 137-30\times 7=289$$$$\Rightarrow x=\sqrt{289}=17$$

Commented by mr W last updated on 14/Nov/22

$$a=\frac{11+4}{\sqrt{137}}\times 11=\frac{165}{\sqrt{137}}$$$$b=\frac{11+4}{\sqrt{137}}\times 4=\frac{60}{\sqrt{137}}$$$$x=\sqrt{{(\sqrt{137}+b)}^2+{(a-\sqrt{137})}^2}$$$$=\sqrt{a^2+b^2+2\times 137+2(b-a)\sqrt{137}}$$$$=\sqrt{{15}^2+2\times 137-2(165-60)}$$$$=17$$

Answered by manxsol last updated on 16/Nov/22

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