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An isosceles triangle of wood of base 2a and height h is placed with its base on the ground vertex directly above. The triangle faces the sun whose altitude is 30^@. Then, the tangent of the angle at the apex of the shadow is |
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Answer» `(2ah)/sqrt3` Let BCE be the shadow of `triangleABC` such that E is the apex of the shadow. Clearly, DE is the height of the TRIANGULAR shadow. Let `2alpha` be the angle at the apex E of the shadow. Then, `tan ALPHA =a/(DE)`  Since the altitude of the sun is `30^@` . Therefore, `tan 30^@=(AD)/(DE)RARR 1/sqrt3=h/(DE) rArr DE=sqrt3h` `therefore tan alpha=a/(sqrt3h)` Now, `tan 2alpha=(2 tan alpha )/(1-tan^2 alpha ) rArr tan 2alpha =(2a//sqrt3h)/(1-a^2/(3h^2))=(2sqrt3ah)/(3h^2-a^2)` |
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