3. A man 1.5 m tall observes that the angle ofelevation of the top of

1.	3. A man 1.5 m tall observes that the angle ofelevation of the top of a building is 60°. Ifhe stands 50 m away from the base of thebuilding, find its height to the nearest metre.
Answer» $Height \:of \:a \:man (AB) = 1.5 \:m$ $Height \:of \:the \: building (CE) = CD + DE \\= (1.5m + x) \:m$ $Distance \: between \: man \: to \:base \\of \: of \:the \: building (BC) = 50 \:m$ $Angle \:of \: elevation = 60\degree$ $In \:triangle ADE, \:we \:have , \\tan \:60\degree = \frac{ED}{AD} \\\implies \sqrt{3} = \frac{x}{50} \\\implies 50\sqrt{3} = x$ $Height \: of \:the \: building = x + CD \\= 50\sqrt{3} + 1.5 \\= 50\times 1.732 + 1.5 \\= 86.6 + 1.5 \\=88.1 \:m$ THEREFORE., $\red {Height \: of \:the \: building}\green {=88.1 \:m}$ •••♪

3. A man 1.5 m tall observes that the angle ofelevation of the top of a building is 60°. Ifhe stands 50 m away from the base of thebuilding, find its height to the nearest metre.

Answer»

$Height \:of \:a \:man (AB) = 1.5 \:m$

$Height \:of \:the \: building (CE) = CD + DE \\= (1.5m + x) \:m$

$Distance \: between \: man \: to \:base \\of \: of \:the \: building (BC) = 50 \:m$

$Angle \:of \: elevation = 60\degree$

$In \:triangle ADE, \:we \:have , \\tan \:60\degree = \frac{ED}{AD} \\\implies \sqrt{3} = \frac{x}{50} \\\implies 50\sqrt{3} = x$

$Height \: of \:the \: building = x + CD \\= 50\sqrt{3} + 1.5 \\= 50\times 1.732 + 1.5 \\= 86.6 + 1.5 \\=88.1 \:m$

THEREFORE.,

$\red {Height \: of \:the \: building}\green {=88.1 \:m}$

•••♪

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