Question
ABC is an isosceles triangle inscribed in a circle. If
AB = AC = 12√5 and BC = 24 cm then radius of circle isSolution
we know if we draw a Ʇ which AD is bisect BC in two equal part BD = DC = 12 Now apply PGT in ∆ADC AD = 24 If 0 in centre of circle. then OD = (24 – r) Now apply PGT in ∆OBD we get r2 = (12)² + (24 – r)2 r2 – (24 – r )2 = 12² (r + 24 – r )(r – 24 + r) = 144 2r = 24 + 6 = 2r = 30 = r = 15
If cosec (2A + B) = 2 and cosec (A + B) = 1, find the value of (3A – B).
sin2 12˚ + sin2 14˚ + sin2 16˚ + sin2 18˚ + ……… + sin2 78˚ =?
If cot5A = tan(3A-14˚), find the value of A? Given that 5A and 3A-14˚ are acute angles.
If x = 4 cos A + 5 sin A snd y = 4 sin A – 5 cos A, then the value of x2+y2 is :
Solve the following trigonometric expression:
If sec 4A = cosec (3A - 50°), where 4A and 3A are acute angles, find the value of A + 75.
Find the maximum value of 18 sin A + 7 cos A.
If sin(3x - 2o) = cos(2x + 22o), then find the value of: {3tan(4x - 11o) + cosec(5x - 10o)} X cot(2x + 2
Simplify the following trigonometric expression:
10 cos 71° sec 19° − 4 cot 62° tan 28°Simplify: sin (A + B) sin (A - B)
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