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

Right triangle trigonometry Quiz 2

mathematics

 What is $\cos 30{}^\circ $ multiplied by the perimeter of a right triangle with perpendicular side $4$  and base  $3$

 Find the missing side, find the perimeter and multiply with the value of $\cos 30{}^\circ $ to get the answer.

Let the hypotenuse of the given triangle be $x$. $ \\ $ Applying Pythagoras theorem to the given triangle, $ \\ {{x}^{2}}={{4}^{2}}+{{3}^{2}} \\ \Rightarrow x=5 \\ $ So, the perimeter of the given triangle is, $ \\ P=5+4+3=12 \\ \cos \left( 30{}^\circ  \right)=\dfrac{\sqrt{3}}{2} \\ $ So, the required answer is, $\dfrac{\sqrt{3}}{2}\times 12=6\sqrt{3}$

 If AB is a bisector of the isosceles triangle and $\angle C=\angle D=\alpha $ .What is the value of  $2\alpha $ $ \\ $  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/c6770fd4-0492-429c-9083-b5b2d5880a1c114868201487816485.png"/>

 Use the properties of isosceles triangles to find the missing values of the triangle. Use these values in the law of cosines rule and find the desired result.

As $\angle C=\angle D=\alpha $, so the sides opposite the angles are also equal,  $ \\ \therefore AC=AD=\sqrt{2} \\ $ AB is the bisector, so $ \\ CD=1+1=2 \\ $ Applying law of cosines rule for the $\Delta ACD$, $ \\ A{{D}^{2}}=A{{C}^{2}}+C{{D}^{2}}-2(AC)(CD)\cos \left( \angle C \right) \\ \Rightarrow {{\left( \sqrt{2} \right)}^{2}}={{\left( \sqrt{2} \right)}^{2}}+{{2}^{2}}-2\left( \sqrt{2} \right)(2)\cos \left( \alpha  \right) \\  \Rightarrow 4\left( \sqrt{2} \right)\cos \left( \alpha  \right)=6-2 \\ \Rightarrow \cos \left( \alpha  \right)=\dfrac{1}{\sqrt{2}} \\ \Rightarrow \alpha =45{}^\circ  \\ \therefore 2\alpha =90{}^\circ \\ $

 Find the perimeter of a right angled triangle with perpendicular $5\text{ cm}$ and angle opposite to perpendicular is   $\text{30}{}^\circ $. 

 First draw the rough figure to get an idea what we have and what we need. Use trigonometric ratios to find the missing sides. Then get the perimeter.

<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/4e220f3c-1e8b-4d86-8de8-e40fdf496d904642763085788881863.png"/> $ \\ $  It is given that the angle opposite to perpendicular is $30^\circ$, i.e., $\angle C=30{}^\circ $. $ \\ $ We know, $ \\  \sin 30{}^\circ =\dfrac{AB}{AC}=\dfrac{5}{AC} \\ \Rightarrow AC=10cm \\ $ Now applying Pythagoras theorem, we get $ \\ AC=\sqrt{A{{B}^{2}}+B{{C}^{2}}} \\  {{10}^{2}}={{5}^{2}}+B{{C}^{2}} \\  \Rightarrow BC=\sqrt{75}=8.6cm \\ $ So, the perimeter is $ =10+5+8.6=23.6cm$

 If  $\tan X=\dfrac{3}{4}$  .Then find $\dfrac{1}{3}\csc X$.

 Use trigonometric ratios to find all the sides of the given triangle then use the formula of required trigonometric ratio to get the required answer.

<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/b2da3386-7ff7-46ea-a23c-c7ad6874a9233852549901219568012.png"/>  $ \\ $ Now applying Pythagoras theorem, we have, $A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}} \\  \Rightarrow A{{C}^{2}}={{3}^{2}}+{{4}^{2}} \\  \Rightarrow AC=\sqrt{25}=5 \\ $ Now, by using cosecant ratio the required value is $ \\ \Rightarrow \dfrac{1}{3}\csc X=\dfrac{1}{3}\times \dfrac{AC}{AB} \\ \Rightarrow \dfrac{1}{3}\csc X=\dfrac{1}{3}\times \dfrac{5}{3}=\dfrac{5}{9}$

 Find the product of areas of the right-angled triangles if $\angle C=60{}^\circ $ and $\angle D=45{}^\circ $. Given that the sides can be defined by $\text{cosec }60{}^\circ $ and $\sec 45{}^\circ $ for respective triangles. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/cdb88e6e-9cd1-429b-b718-8bf94634c8b83227502815939479863.png"/>

 Use trigonometric ratios to find the sides of the triangles, then apply the formula, $\text{area}=\dfrac{1}{2}\times \text{base}\times \text{height}$. Then multiply the area of the two triangles to get the desired result.

Consider $\Delta ABC$,  $ \\ $ Given its sides can be obtained from $\text{cosec }60{}^\circ $ $ \\ \Rightarrow \csc {{60}^{\circ }}=\dfrac{AC}{AB} \\  \Rightarrow \dfrac{AC}{AB}=\dfrac{2}{\sqrt{3}} \\ $ Therefore,$AC=2,AB=\sqrt{3}$ $ \\ $ Applying Pythagoras, we get $ \\ A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}} \\ \Rightarrow 4=3+B{{C}^{2}} \\ \Rightarrow BC=1 \\ $ Area of $\Delta ABC$ $ \\ {{A}_{1}}=\dfrac{1}{2}\times \text{base}\times \text{height}=\dfrac{1}{2}\times 1\times \sqrt{3}=\dfrac{\sqrt{3}}{2}..........(i) \\ $ Consider $\Delta DEF$ $ \\ $ Given its sides can be obtained from $\sec 45{}^\circ $, $ \\ \Rightarrow \sec {{45}^{\circ }}=\dfrac{DE}{EF} \\ \Rightarrow \dfrac{DE}{EF}=\dfrac{\sqrt{2}}{1} \\ $ Therefore,$DE=\sqrt{2},FE=1$ $ \\ $ Area of $\Delta DEF $  $ \\ {{A}_{2}}=\dfrac{1}{2}\times \text{base}\times \text{height}=\dfrac{1}{2}\times \sqrt{2}\times 1=\dfrac{1}{\sqrt{2}}..........(ii) \\ $ From equation $(i)$ and $(ii)$, product of areas will be $ \\ A=\dfrac{\sqrt{3}}{2}\times \dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{3}}{2\sqrt{2}}$