In this blog on trigonometry, we will solve some trigonometry questions and answers to brush up on our concept using identities.  Trigonometry is a combination of two Greek words- ‘Trigonon’ meaning a triangle and ‘metron’ meaning measure. In this blog, we will solve some of the problem statements to understand this concept better. Test your understanding of the practice problem and solve the step-by-step solution.

Solved Trigonometry questions

Problem 1: Find the length of the side unknown?
Length of the triangle whose side is unknown-trigonometry questions and answers
Solution :

x = \frac{20} {tan(45°)} = 20 

H = \frac{20} {sin(45°)} = 28.28

Problem 2:Find all sides of the right-angle triangle if the area of the is 625.
Evaluate sides of right angle triangle when area is known

Here, area of a triangle =\frac{1}{2}base *height

Area of triangle=\frac{1}{2} (2a)(a) = 625

So, a = 25 , 2a = 50

Using Pythagoras’s theorem
(opp.)^2 + (adj.)^2 = (Hypo.)^2 \\

(2a)^2 + (a)^2 = H^2

H = a √(5)= 25 √(5)

Problem 3:Find the length if DB is perpendicular to AB.
Length of unknown side
Solution :

Given AC perpendicular to DB means that triangles ACD and ACB are right triangles.
\ tan (30°) = \frac {10} {AC} or AC = \frac {10} {tan (30°)}

BC = 20 – AC = 19 - \frac {10} {tan (30°)}

Pythagoras’s theorem applied to right \triangle ACB:
11^2 + BC^2 = x^2

solve for x and substitute BC: x = \sqrt ([ 10^2 + (20 - \frac {10} {tan (30°)})^2 )

= 107.12 (rounded to 3 significant digits)

Problem 4: Find the value of \tan \theta Evaluating side of triangle with known angle
Solution :

We have the opposite side (O) =10 and Adjacent side = x

\tan \theta =\frac{opposite}{adjacent}

\tan 30= \frac{10}{x}

x=\frac{10}{\ tan 30}

x= 17.33

Problem 5: Solve 2 \tan(\theta)- \cot(\theta)=-1

Solution :

2\tan\theta-\cot \theta=-1  \\

2\tan^2\theta+tan\ \theta-1=0 


Now,tan\theta+1=0 or 2tan\theta-1=0

tan\theta=-1=tan\left(\frac{-\pi}{4}\right)or tan\theta=\frac{1}{2\ }=tan\alpha\left(say\right)\\

\theta=n\pi+(\frac{-\pi}{4})\ or\ \theta=n\pi+\alpha,              Where \alpha=\tan^{-1}(\frac{1}{2}) and\ n\in Z

Problem 6: Solve \tan 2 \theta \tan \theta = 1


tan\left(2\theta\right)tan\ \theta=1 

\left(\frac{sin\ 2\theta}{cos\ 2\theta}\right)\left(\frac{sin\ \theta}{cos\ \theta}\right)=1 


cos\ 2\theta cos\ \theta-sin\ 2\theta sin\ \theta=0 cos\left(2\theta+\theta\right)=0 

cos\ 3\theta=0 


\theta=\left(2n+1\right)\frac{\pi}{6}\ \ or\ \theta=\left(\frac{n\pi}{3}+\frac{\pi}{6}\right)

Problem 7:  Solve \tan (\frac{\pi}{4}+\theta)+\tan (\frac{\pi}{4}-\theta)=4


\frac{1+tan\ \theta}{1-tan\ \theta}+\frac{1-tan\ \theta}{1+tan\ \theta}=4 

(1+tan\ \theta)^2+\ (1+tan\ \theta)^2=4(1+tan\ \theta)(1-tan\ \theta)

2+2\ tan^2\theta=4(1+tan\ \theta\ -tan\theta-tan^2\ \theta

2+2\ tan^2\theta=4-4\ tan^2\theta6\ tan^2\theta=2\tan^2\theta=\frac{1}{3}=\ (\frac{1}{\sqrt3})^2=tan^2\frac{\pi}{6}

\theta=n\pi\pm\frac{\pi}{6},\ where\ n\in Z

Problem 8: In \triangle PQR, Prove that 4[bc \cos^2 (\frac{P}{2})+ca \cos^2 (\frac{Q}{2})+ab cos^2 (\frac{R}{2})=(a+b+c)^2


L.H.S =

=4[bc(\frac{1+cos\ P}{2})+ca(\frac{1+cos\ Q}{2})+ab(\frac{1+cos\ R}{2}) 

=2\left(bc+ab+bc\right)+2\left(bc c o sP+ca c o sQ+ab c o sR\right)



Problem 9: In \triangle PQR, Prove that \frac{1+\cos(P-Q)\cos R}{1+\cos(P-R)\cos Q}=\frac{a^2+b^2}{a^2+c^2}



cos\ Q\ =\ \ cos\ 180°-P+Q=-cosP+Q

=\frac{1+cos\left(P-Q\right)-cos\left(P+Q\right)}{1+cos\ \left(P+R\right)\left\{-cos\left(P+R\right)\right\}}=\frac{1-\left(cos^2P-sin^2Q\right)}{1-\left(cos^2P-sin^2R\right)}




Problem 10: Evaluate \tan (\frac{19\pi}{3})



=tan\ \frac{\pi}{3}


Problem 11: Prove that: Prove that \sin^26x -\sin^2 4x=\sin 2x \sin 10x





Using identity : \sin^2X-\sin^2Y=sin(X+Y)sin(X-Y)

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