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To find the distances and heights we can use the mathematical techniques, which come under the Trigonometry. It shows the relationship between the sides and the angles of the triangle. Generally, it is used in the case of a right angle triangle.
In a right angle triangle, the ratio of its side and the acute angles is the trigonometric ratios of the angles.
In this right angle triangle ∠B = 90°. If we take ∠A as acute angle then -
AB is the base, as the side adjacent to the acute angle.
BC is the perpendicular, as the side opposite to the acute angle.
Ac is the hypotenuse, as the side opposite to the right angle.
Trigonometric ratios with respect to ∠A
Remark
Cosec A, sec A, and cot A are the reciprocals of sin A, cos A, and tan A respectively.
uotient Relation
Example
Find the lengths of the sides BC and AC in ∆ ABC, right-angled at B where AB = 25 cm and ∠ACB = 30°, using trigonometric ratios.
Solution
To find the length of the side BC, we need to choose the ratio having BC and the given side AB. As we can see that BC is the side adjacent to angle C and AB is the side opposite to angle C, therefore
tan 30°
BC = 25√3 cm
To find the length of the side AC, we consider
AC = 50 cm
If the sum of two angles is 90° then, it is called Complementary Angles. In a right-angled triangle, one angle is 90 °, so the sum of the other two angles is also 90° or they are complementary angles.so the trigonometric ratios of the complementary angles will be -
sin (90° – A) = cos A,
cos (90° – A) = sin A,
tan (90° – A) = cot A,
cot (90° – A) = tan A,
sec (90° – A) = cosec A,
cosec (90° – A) = sec A
An equation is said to be a trigonometric identity if it contains trigonometric ratios of an angle and satisfies it for all values of the given trigonometric ratios.
In ∆PQR, right angled at Q, we can say that
PQ2 + QR2 = PR2
Divide each term by PR2, we get
(sin R)2 + (cos R)2 = 1
sin2 R + cos2 R =1
Likewise other trigonometric identities can also be proved. So the identities are-
sin2 R + cos2 R = 1
1 + tan2 R = sec2 R
cot2 R + 1 = cosec2 R
How to solve the problems related to trigonometric ratios and identities?
Prove that
L.H.S
(by reciprocal identity and quotient identity)
(by simplifying the denominator and by Pythagoras identity)
Hence, L.H.S = R.H.S
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