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In this lesson, we will learn how to solve equations by using inverse trigonometric functions.

Q1:

If πΈ is an acute angle. Find the value of π β πΈ given s i n πΈ = 0 . 7 9 1 . Give the answer to the nearest second.

Q2:

Find the set of values satisfying s i n π₯ = β β 2 2 , where 0 β€ π₯ < 2 π .

Q3:

Find the set of values satisfying t a n π₯ = β 1 β 3 , where 0 β€ π₯ < 2 π .

Q4:

Given that s i n π΄ = 0 . 8 1 9 3 , determine π β π΄ to the nearest tenth of a degree.

Q5:

Find the set of values satisfying c o s π₯ = 1 2 , where 0 β€ π₯ < 2 π .

Q6:

Find the set of values satisfying t a n π = 0 given 0 < π < 3 6 0 β β .

Q7:

Find the measure of β π in degrees given 2 π = 6 0 c o s t a n β where π is an acute angle.

Q8:

Find the measure of β π in degrees given t a n t a n π = β 3 4 5 β where π is an acute angle.

Q9:

Find, to the nearest tenth of a degree, the size of the angle of c o s π΄ = 0 . 6 1 9 4 .

Q10:

Find the smallest positive angle that satisfies both 2 π β β 2 = 0 c o s and t a n π β 1 = 0 .

Q11:

Find the measure of β π΄ given 1 7 π΄ β 1 6 = 0 s i n where π΄ β ο 0 , π 2 ο . Give the answer to the nearest second.

Q12:

Find the angle π΄ to the nearest tenth of a degree, knowing that t a n π΄ = 0 . 8 6 and π΄ β ] 0 , 1 8 0 [ β β .

Q13:

Find π β πΈ given t a n πΈ = 1 8 . 5 8 4 5 and β πΈ is an acute angle. Give the answer to the nearest second.

Q14:

Find the value of π β πΈ given that β πΈ is an acute angle and c o s πΈ = 0 . 5 2 0 1 . Give the answer to the nearest second.

Q15:

Find the value of π β πΈ given that β πΈ is an acute angle and c o s πΈ = 0 . 7 7 9 . Give the answer to the nearest second.

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