Solution of a Right Angled Triangle
Finding the solution of right angled triangle means to find the all six parameters of the triangle. In the following cases solution of a triangle can be found:
If three elements of any triangle are known, then the remaining elements can be calculated.
In a right-angled triangle, one element is known i.e. one right angle.
So a right-angled triangle can be solved if
two sides of the triangle are given.
one side and one acute angle are given.
Cosecant (cosec) of an angle: The ratio of the hypotenuse to the perpendicular is called the cosecant of the angle.
Cosine (cos) of an angle: The ratio of the base or adjacent side to the hypotenuse is called the cosine of the angle.
Cotangent (cot) of an angle: The ratio of the base to the perpendicular is called the cotangent of the angle.
Secant (sec) of an angle: The ratio of the hypotenuse to the base is called the secant of the angle.
Sine (sin) of an angle: The ratio of the perpendicular or opposite side to the hypotenuse is called the sine of the angle.
Tangent (tan) of an angle: The ratio of the perpendicular to the base is called the tangent of the angle.
Trigonometric ratio: The ratio between the lengths of a pair of two sides of a right-angled triangle is called trigonometric ratio.
Trigonometric Ratios of Complementary Angles: If the sum of two angles is 90°, then they are said to be complementary angles.
Sin (90° – ?) = cos ?
cos (90° – ?) = sin ?
tan (90° – ?) = cot ?
cosec (90° – ?) = sec ?
sec (90° – ?) = cosec?
cot (90° – ?) = tan ?
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If cos9A= sinA and 9AMarks:2
Since, cos9A = sinA, then
cos9A = cos (90 - A)
9A = 90 - A
10A = 90
A = 90/10
So, we have
tan5A = tan5(9)
If sin3A = cos (A – 26), where 3A is an acute angle, find the value of A.Marks:2
Since, sin3A = cos (A – 26)
cos (90 - 3A) = cos (A – 26)
90 - 3A = A – 26
90 +26= A + 3A
116 = 4A
A = 116/4
Evaluate: sin 13 - cos 87Marks:1
sin 13 - con 87
= cos (90- 13) - cos 87
[Since, sin x = cos(90- x) ]
= cos 87 - cos 87
If tan A = cot B, prove that A + B = 90.Marks:1
Given,tan A = cot B
tan A = tan (90 − B) ( Since, tan (90 − x) = cot x )
or, A = 90 − B
or, A + B = 90
Show that:tan48 tan23 tan42 tan67 = 1Marks:1
tan 48 tan 23 tan 42 tan 67
= tan (90 − 42) tan (90 − 67) tan 42 tan 67
= cot 42 cot 67 tan 42 tan 67
= (cot 42 tan 42) (cot 67 tan 67)
= (1) (1)