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Question Number 89290 by I want to learn more last updated on 16/Apr/20

Commented by jagoll last updated on 16/Apr/20

$$diagonal=\sqrt{{(14+9)}^2+7^2}$$$$=\sqrt{{23}^2+49}=\sqrt{578}=17\sqrt{2}cm$$$$thenx=17cm$$

Commented by I want to learn more last updated on 16/Apr/20

$$\mathrm{I}\mathrm{appreciate}\mathrm{sir}$$

Answered by TANMAY PANACEA. last updated on 16/Apr/20

$$tan\theta =\frac{a}{9}=\frac{7-a}{14}$$$$14a=63-9a$$$$a=\frac{63}{23}$$$$tan\theta =\frac{63}{23\times 9}=\frac{7}{23}$$$$sin\theta =\frac{7}{\sqrt{{23}^2+7^2}}=\frac{7}{\sqrt{529+49}}=\frac{7}{\sqrt{578}}$$$$sin\theta =\frac{a}{l_1}{l}_1=\frac{a}{sin\theta }$$$$sin\theta =\frac{7-a}{l_2}$$$$l_2=\frac{7-a}{sin\theta }$$$$l_1+l_2=diagonal=x\sqrt{2}$$$$x\sqrt{2}=\frac{7}{sin\theta }=\frac{7}{7}\times \sqrt{578}$$$$x=\sqrt{289}=17$$

Commented by I want to learn more last updated on 16/Apr/20

$$\mathrm{I}\mathrm{appreciate}\mathrm{sir}$$

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