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This is one of our many quizzes on Trigonometry vocabulary. Click on this to access more such competitive quizzes on Trigonometry.

Trigonometry vocabulary Quiz 2

mathematics

 A man is walking towards a current pole which is perpendicular to the ground. He stopped at a distance of $8mts$ to the pole and he can see the top of the pole with an angle ${{45}^{\circ }}$. Then find the length of the pole?

 Conceptualize the given problem and then use the appropriate trigonometric ratio. Using tan is better for this problem since adjacent is given for finding hypotenuse.

  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/4e57ce6c-f33a-4baa-85bb-667e8d2c0add4395668339453539412.png"/> $ \\  \text{Using tan}\theta \text{ } \\  \text{tan4}{{\text{5}}^{\circ}}=\dfrac{\text{AC}}{\text{CB}} \\  \Rightarrow 1=\dfrac{\text{AC}}{\text{CB}} \\  \Rightarrow \text{AC=8} \\  \Rightarrow \text{h=8} \\ $

 A helicopter is approaching point X along a straight line and maintaining constant altitude A. At 9:00 am, the angle of elevation of the helicopter is ${{30}^{\circ }}$ and at 9:01 am it is ${{60}^{\circ }}$. What is the altitude A of the helicopter if the speed of the helicopter is constant and equal to 300 miles/hour?

 Imagine the concept and draw a diagram based on it. Using the Tan trigonometric ratio find the two equations and then solve them to get the altitude $(A)$.

  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/0cf7c3d4-5fcf-4afa-a38c-78ccda99872f4376197419858761791.png"/> $ \\  \text{Distance travelled in one minute d = }\dfrac{300}{60}=5 \\  \text{Using tan } \\  \tan {{30}^{\circ}}=\dfrac{\text{A}}{5+x}........\left( 1 \right) \\  \tan {{60}^{\circ }}=\dfrac{\text{A}}{\text{X}} \\  \Rightarrow \sqrt{3}=\dfrac{\text{A}}{\text{X}} \\  \Rightarrow \text{X=}\dfrac{\text{A}}{\sqrt{3}} \\  \text{Substituting in the equation (1)} \\  \Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{\text{A}}{5+\text{X}} \\  \Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{\text{A}}{5+\dfrac{\text{A}}{\sqrt{3}}} \\  \text{By solving the above equation we get } \\  \Rightarrow \text{A=}\dfrac{5\sqrt{3}}{2} \\ $

 $\text{Find }\cos \left( {{34}^{\circ}} \right)+\cos \left( {{6}^{\circ}} \right)+\cos \left( {{174}^{\circ}} \right)+\cos \left( {{214}^{\circ}} \right)+\cos \left( {{0}^{\circ}} \right)\text{ ?}$

 Evaluate the given cos angles using ASTC rule (All silver tea cups rule) and then verify the options.

$ \text{Writing terms in their suitable way}\text{.} \\  \Rightarrow \cos \left( 34 \right)+\cos \left( 6 \right)+\cos \left( 180-6 \right)+\cos \left( 180+34 \right)+1 \\  \Rightarrow \cos \left( 34 \right)-\cos \left( 34 \right)+\cos \left( 6 \right)-\cos \left( 6 \right)+\cos \left( 0 \right) \\  \Rightarrow 1 \\  \text{Checking each options, option (c) is correct} \\  2\left( {{\cos }^{2}}\left( 23 \right)+{{\sin }^{2}}\left( 23 \right) \right)-{{\sec }^{2}}\left( 89 \right)+{{\tan }^{2}}\left( 89 \right) \\  \Rightarrow 2-1 \\  \Rightarrow 1 \\ $

 Evaluate $\dfrac{\sin {{15}^{\circ}}}{\cos {{15}^{\circ}}}$?

 Apply the formulas $\sin \left( \text{A-B} \right)$ and $\text{cos}\left( \text{A-B} \right)$ to the equation and then substitute the values to get solved.

 $\text{Writing the given equation as }\dfrac{\sin \left( {{45}^{\circ}}-{{30}^{\circ}} \right)}{\cos \left( {{45}^{\circ}}-{{30}^{\circ}} \right)} \\  \text{By the formulas } \\  \Rightarrow \sin \left( \text{A-B} \right)=\sin \text{AcosB-cosAsinB} \\  \Rightarrow \text{cos}\left( \text{A-B} \right)=\cos \text{AcosB+sinAsinB} \\  \text{By applying this formulas } \\  \Rightarrow \dfrac{\sin \left( {{45}^{\circ}}-{{30}^{\circ}} \right)}{\cos \left( {{45}^{\circ}}-{{30}^{\circ}} \right)}=\dfrac{\sin {{45}^{\circ}}\text{cos3}{{\text{0}}^{\circ}}\text{-cos4}{{\text{5}}^{\circ}}\text{sin3}{{\text{0}}^{\circ}}}{\text{cos4}{{\text{5}}^{\circ}}\text{cos3}{{\text{0}}^{\circ}}\text{+sin4}{{\text{5}}^{\circ}}\text{sin3}{{\text{0}}^{\circ}}} \\  \Rightarrow \dfrac{\left( \dfrac{1}{\sqrt{2}} \right)\left( \dfrac{\sqrt{3}}{2} \right)-\left( \dfrac{1}{\sqrt{2}} \right)\left( \dfrac{1}{2} \right)}{\left( \dfrac{1}{\sqrt{2}} \right)\left( \dfrac{\sqrt{3}}{2} \right)+\left( \dfrac{1}{\sqrt{2}} \right)\left( \dfrac{1}{2} \right)} \\  \Rightarrow \dfrac{\sqrt{2}\left( \sqrt{3}-1 \right)}{\left( \sqrt{6}+\sqrt{2} \right)} \\ $

 Find the general solution for ${{\cos }^{2}}x+\cos 2x=\dfrac{1}{2}$?

 Take $\dfrac{1}{2}$ on LHS and take LCM then convert $2\cos ^2\left(x\right)+2\cos \left(2x\right)-1=3\cos \left(2x\right)$ solve the equation and write its general form.

 $\cos ^2\left(x\right)+\cos \left(2x\right)-\dfrac{1}{2}=0 \\  \dfrac{2\cos ^2\left(x\right)+2\cos \left(2x\right)-1}{2}=0 \\ 2\cos ^2\left(x\right)+2\cos \left(2x\right)-1=0 \\ 3\cos \left(2x\right)=0 \\ 2x=\dfrac{\pi }{2}+2\pi n,\:2x=\dfrac{3\pi }{2}+2\pi n \\ x=\dfrac{\pi }{4}+\pi n,\:x=\dfrac{3\pi }{4}+\pi n $