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In the figure BC = 12 ∠D = 90°,Find ∠CBD, ∠ACD, ∠ABC . Find BD, CD, AD, AC, AB. Find the ratio of the sides of the triangle having the angles 30°, 15°, 135° . |
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Answer» ∠ACD = 180 – (30 + 90) = 60 ∠BCD = 60 – 15 = 45 ∠CBD = 45 ∠ABC = 135 [∵ 180 – (30 + 15)] BD = \(\frac{BC}{\sqrt{2}}=\frac{12}{\sqrt{2}}=6\sqrt{2}\) = 8.48 CD = \(6\sqrt{2}\) (∵ ∆BCD is an isosceles right angles) AC = 2 x \(6\sqrt{2}=12\sqrt{2}\) AD = \(6\sqrt{2}\times\sqrt{3}=6\sqrt{6}=14.69\) AB = AD - BD = 14.69 - 8.48 = 6.21 The ratio of the sides of the triangle with angles 30°, 15°, 135° are 12 : 6,21 : \(12\sqrt{2}\) |
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