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Question: A vertical tower stands on a horizontal plane and is surmounted by a flag-staff of height 7m from a ...

A vertical tower stands on a horizontal plane and is surmounted by a flag-staff of height 7m from a point on the plane, the angle of elevation of the bottom and the top of flag-staff are 300,450{{30}^{0}},{{45}^{0}} respectively. Find the height of the tower.

Explanation

Solution

Hint:To solve the question, we have to apply trigonometry rules for the figure drawn from interpreting the given information.

Complete step-by-step Solution:

Given

The height of the flag-staff surmounted on vertical tower = 7 metres

Let AB be the height of the flag-staff and BC be the height of the vertical tower and α,β\alpha ,\beta be the angle of elevation of the bottom and the top of flag-staff respectively.

AB = 7 metres and AC = AB + BC = (7 + BC) metres.

The given angle of elevation of the bottom of the flag-staff = 300{{30}^{0}}

The angle between BD and AC is equal to 300{{30}^{0}}

α=300\Rightarrow \alpha ={{30}^{0}}

The given angle of elevation of the top of the flag-staff = 450{{45}^{0}}

β=450\Rightarrow \beta ={{45}^{0}}

The angle between AD and AC is equal to 450{{45}^{0}}

By applying trigonometry of angles, we get

tanα=BCCD\tan \alpha =\dfrac{BC}{CD}

tanβ=ACCD\tan \beta =\dfrac{AC}{CD}

By substituting the angles and values in the above formula we get,

tan300=BCCD\tan {{30}^{0}}=\dfrac{BC}{CD}

tan450=7+BCCD\tan {{45}^{0}}=\dfrac{7+BC}{CD}

We know that the values of tan450\tan {{45}^{0}}is equal to 1 and the value of tan300\tan {{30}^{0}}is equal to 13\dfrac{1}{\sqrt{3}}.

13=AC7CD\Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{AC-7}{CD} and 1=ACCD1=\dfrac{AC}{CD}

CD=3BC\Rightarrow CD=\sqrt{3}BC and CD=7+BCCD=7+BC

By solving the above expression, we get

3BC=7+BC\sqrt{3}BC=7+BC

BC(31)=7BC\left( \sqrt{3}-1 \right)=7

BC=7(31)=71.731=70.73=9.58BC=\dfrac{7}{\left( \sqrt{3}-1 \right)}=\dfrac{7}{1.73-1}=\dfrac{7}{0.73}=9.58metres

Thus, the height of the tower is equal to 9.58 metres.

Note: The possibility of mistake can be not applying trigonometry for solving the problem. The alternative way for easing the procedure is by using other cosine, sine, cot angles for solving. The other alternative quick way of solving is applying the direct formula for the height of the tower xtanα(tanβtanα)\dfrac{x\tan \alpha }{\left( \tan \beta -\tan \alpha \right)} where x, are height of the flag-staff, the angle of elevation of the bottom and the top of flag-staff respectively.