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This is one of our many quizzes on Angles of elevation and depression. Click on this to access more such competitive quizzes on Trigonometry.

Angles of elevation and depression Quiz 2

mathematics

 A man measures the angle of elevation between the ground and a building $40m$ away to be ${{30}^{0}}$ . How tall is the building?

 Use the formula for tangent of an angle and find out the value of height of the tower from that.

Given, the angle of elevation ${{30}^{0}}$. $ \\ $ We know that, $\tan \theta =\dfrac{opposite\,side}{adjacent\,side}$ $ \\ $  Let the height of tower be $h$ $ \\ \Rightarrow \tan {{30}^{0}}=\dfrac{h}{40} \\ \Rightarrow h=40\times \left( \dfrac{1}{\sqrt{3}} \right) \\ \therefore h=\dfrac{40}{\sqrt{3}}m$

 From an observation point on the ground, the angle of elevation to the top of a very tall tree measures ${{50}^{0}}$. We move 10ft further away from the tree and measure again the angle of elevation to the top of the tree, which now shows ${{27}^{0}}$. How tall is the tree?

 First draw the appropriate diagram for the given information and then find necessary values by using trigonometric ratios to find the height of the tree.

<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/7c7969d6-992a-487b-94a0-0a8a6662292f1010092466710147684.png"/> $ \\ $   Let assume the height of the tree as $h$ $ \\ $  Now consider $\Delta CTA$, here $\tan {{50}^{0}}=\dfrac{h}{x} \\ \Rightarrow x=\dfrac{h}{\tan {{50}^{0}}}$ $ \\ $  Now consider $\Delta CTB$, here ,$\tan {{27}^{0}}=\dfrac{h}{x+10} \\ \Rightarrow x+10=\dfrac{h}{\tan {{27}^{0}}}$ $ \\ $  Hence, $\dfrac{h}{\tan {{27}^{0}}}-\dfrac{h}{\tan {{50}^{0}}}=10 \\\Rightarrow h=10\left( \dfrac{1}{\tan {{27}^{0}}}-\dfrac{1}{\tan {{50}^{0}}} \right)$ $ \\ $ $\therefore h=8.9ft$

 Rohith wants to watch a match that is being played in the stadium. He can watch the match only if he can climb the nearest tower, so as to climb the tower he uses the wooden boxes of height $20\sqrt{3}$. The stadium is at a distance of $300\,m\,$ from the tower and the angle of elevation to the tower from the stadium is ${{30}^{0}}$. Find the number of boxes used by Rohith to watch the match.

 First find the height of the tower , and then  find the number of boxes used by using the height of each box .

 Let the height of the tower be $h$ $ \\ $ Given, the angle of elevation ${{30}^{0}}$. $ \\ $ We know that, $\tan \theta =\dfrac{opposite\,side}{adjacent\,side} \\ \Rightarrow \tan {{30}^{^{0}}}=\dfrac{h}{300} \\ \Rightarrow h=300\times \dfrac{1}{\sqrt{3}}\,\,=\,100\sqrt{3} \\ $ Given, the height of each box is  $20\sqrt{3}$. $ \\ $ The number of boxes used are $\dfrac{h}{20\sqrt{3}}$ $ \\ \Rightarrow \dfrac{100\sqrt{3}}{20\sqrt{3}}=5$

 The angle of depression of vertex C from vertex A is ${{42}^{0}}$. Calculate the angle of elevation of vertex C from B. Given AB=25m and BD=10m. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/91480efc-eae8-48b6-b2d1-7371df9fd2a53463388275148769213.png"/>

 Divide the given figure into $\Delta AOC$, then find  $AO$ to find $DC$and then by using tan ratio find the required angle.

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/4b46e67c-f3b6-4e48-a542-4fe1b8ac72854916125522813599991.png"/> $ \\ $ Let us first draw a line from C, parallel to AE and meet at AB and name the point as O. $ \\ $  Now consider, $\Delta AOC$ here we know that the $\angle ACO={{42}^{0}}$ $ \\ $  Now, $\tan {{42}^{0}}=\dfrac{AO}{OC}=\dfrac{h}{10}$, we get  $AO=h=9m$ $ \\ $  By this, $OB=DC=16m$ $ \\ $  Now consider, $\Delta BCD$  let the $\angle CBD=\theta $and $\tan \theta =\dfrac{DC}{BD}=\dfrac{16}{10}$ $ \\ \therefore \theta ={{58}^{0}}$

 A person on ship sees a lighthouse N.W. of himself. After sailing for 12km in a direction ${{15}^{0}}$ south of W, the lighthouse is seen due N. Find the distance of lighthouse from the ship in each position 

 Draw the appropriate diagram and find the distance of the lighthouse from original position and then finally find the  distance of the lighthouse from the ship.

<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/d105c0d6-84f2-4840-84c1-63f1cc5fc0ea4061242780052788184.png"/>     $ \\ $ Here $OQ=12km$ and $\beta ={{15}^{0}}$ $ \\ $ O is the original position of the ship and Q the latter, P is the Lighthouse $ \\ $ Now, $OL=12\cos {{15}^{0}}\,and\,LQ=12\sin {{15}^{0}}$ $ \\ $ Again, $\dfrac{OP}{OL}=\sec {{45}^{0}}$ $ \\ OP=12\cos {{15}^{0}}\sec {{45}^{0}} \\ =12\dfrac{\sqrt{3}+1}{2\sqrt{2}}\sqrt{2} \\ OP=16.392kms \\ PL=OL\tan {{45}^{0}}=OL=12\cos {{15}^{0}} \\ PQ=PL+LQ=OL+LQ \\ =12\cos {{15}^{0}}+12\sin {{15}^{0}} \\ =12\left( \dfrac{\sqrt{3}-1}{2\sqrt{2}}+\dfrac{\sqrt{3}+1}{2\sqrt{2}} \right) \\ \therefore PQ=6\sqrt{6}\,kms$