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ICSE Class 10 Mathematics Question 24 of 28

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Question

Question 73

An inclined plane AC is prepared with its base AB which is √3 times its vertical height BC. The length of the inclined plane is 15 m. Find:

(a) value of θ.

(b) length of its base AB, in nearest metre.

An inclined plane AC is prepared with its base AB which is √3 times its vertical height BC. The length of the inclined plane is 15 m. Find: Maths Competency Focused Practice Questions Class 10 Solutions.
Answer

(a) According to question,

⇒ AB = 3×BC\sqrt{3} \times BC

⇒ AB = 3BC\sqrt{3}BC

⇒ BC = AB3\dfrac{AB}{\sqrt{3}}

From figure,

⇒ tan θ = BCAB\dfrac{BC}{AB}

⇒ tan θ = BC3BC\dfrac{BC}{\sqrt{3}BC}

⇒ tan θ = 13\dfrac{1}{\sqrt{3}}

⇒ tan θ = tan 30°

⇒ θ = 30°.

Hence, θ = 30°.

(b) In right angle triangle ABC,

By pythagoras theorem,

⇒ AC2 = BC2 + AB2

⇒ 152 = (AB3)2\Big(\dfrac{AB}{\sqrt{3}}\Big)^2 + AB2

⇒ 225 = AB23\dfrac{AB^2}{3} + AB2

⇒ 225 = AB2+3AB23\dfrac{AB^2 + 3AB^2}{3}

⇒ 225 = 4AB23\dfrac{4AB^2}{3}

⇒ AB2 = 225×34\dfrac{225 \times 3}{4}

⇒ AB2 = 6754\dfrac{675}{4}

⇒ AB2 = 168.75

⇒ AB = 168.75\sqrt{168.75}

⇒ AB = 12.99 m

Rounding off,

AB = 13 m.

Hence, AB = 13 m.