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In the adjoining figure, ABC is a triangle in which ∠B = 45° and ∠C = 60°. If AD ⊥ BC and BC = 8m, find the length of the altitude AD.

In the figure, ABC is a triangle in which ∠B = 45° and ∠C = 60°. If AD ⊥ BC and BC = 8m, find the length of the altitude AD. Trigonometrical Ratios of Standard Angles, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Trigonometrical Ratios

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Answer

In △ABD,

⇒ tan 45° = ADBD\dfrac{AD}{BD}

⇒ 1 = ADBD\dfrac{AD}{BD}

⇒ BD = AD.

In △ADC,

⇒ tan 60° = ADDC\dfrac{AD}{DC}

⇒ 3=ADDC\sqrt{3} = \dfrac{AD}{DC}

⇒ DC = AD3\dfrac{AD}{\sqrt{3}}

From figure,

BC = BD + DC

⇒8=AD+AD3⇒8=3AD+AD3⇒AD(3+1)3=8⇒AD=833+1\Rightarrow 8 = AD + \dfrac{AD}{\sqrt{3}} \\[1em] \Rightarrow 8 = \dfrac{\sqrt{3}AD + AD}{\sqrt{3}} \\[1em] \Rightarrow \dfrac{AD(\sqrt{3} + 1)}{\sqrt{3}} = 8 \\[1em] \Rightarrow AD = \dfrac{8\sqrt{3}}{\sqrt{3} + 1}

Multiplying numerator and denominator by (3−1)(\sqrt{3} - 1)

⇒AD=833+1×3−13−1⇒AD=8(3−3)(3)2−(1)2 [∵a2−b2=(a+b)(a−b)]⇒AD=8(3−3)3−1⇒AD=8(3−3)2⇒AD=4(3−3) m\Rightarrow AD = \dfrac{8\sqrt{3}}{\sqrt{3} + 1} \times \dfrac{\sqrt{3} - 1}{\sqrt{3} - 1} \\[1em] \Rightarrow AD = \dfrac{8(3 - \sqrt{3})}{(\sqrt{3})^2 - (1)^2} \space [\because a^2 - b^2 = (a+b)(a-b)] \\[1em] \Rightarrow AD = \dfrac{8(3 - \sqrt{3})}{3 - 1} \\[1em] \Rightarrow AD = \dfrac{8(3 - \sqrt{3})}{2}\\[1em] \Rightarrow AD = 4(3 - \sqrt{3}) \text{ m}

Hence, AD = 4(3−3)4(3 - \sqrt{3}) m.

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