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Category: Algebra

Question-190947

Question Number 190947 by Shrinava last updated on 15/Apr/23 Answered by mr W last updated on 15/Apr/23 BDAB=sinβsin30(i)ABAC=sin(β+45)sin45(ii)(i)×(ii):$$\mathrm{1}=\frac{\mathrm{sin}\:\beta\:\mathrm{sin}\:\left(\beta+\mathrm{45}\right)}{\mathrm{sin}\:\mathrm{30}\:\mathrm{sin}\:\mathrm{45}} \