if a triangle with sides 12 unit and 5 unit is inscribed in a circle with BC as diameter, then find the value of sin square theta +cos square theta -1
Is this supposed to be a trick question?
sin^2 Ø + cos^2 Ø - 1
= 1 - 1
= 0
the fact that we have a triangle of sides 12 and 5 in a circle has nothing to do with it.
btw, as an aside to your question ...
Any angle subtended by a diameter is a right-angle, so you would have the 5-12-13 right-angled triangle.
the larger of the acute angles is arctan(12/5)
and the smaller is arctan(5/12) ---> appr 22.6°
note:
sin^2 22.6° + cos^2 22.6° - 1
= 1-1 = 0
To find the value of sin^2(theta) + cos^2(theta) - 1, where theta is the angle in the triangle inscribed in the circle, we need to use trigonometric identities.
1. Draw the triangle ABC with sides 12 units and 5 units, where BC is the diameter of the circle. Let angle BAC be represented as theta (θ).
2. Since BC is the diameter of the circle, angle BDC is a right angle (90 degrees) and can be represented as pi/2 radians.
3. Apply the Pythagorean theorem in triangle BCD, where BD is the radius (half of the diameter) and CD is the remaining length of side BC.
BD^2 + CD^2 = BC^2
(BC/2)^2 + CD^2 = BC^2
(BC/2)^2 + CD^2 = (BC)^2
Substitute BC = 12, as given:
(12/2)^2 + CD^2 = 12^2
6^2 + CD^2 = 12^2
36 + CD^2 = 144
CD^2 = 144 - 36
CD^2 = 108
4. Use the trigonometric identity for sine and cosine:
sin^2(theta) + cos^2(theta) = 1
Therefore, we need to find the ratio of CD (opposite side) to BD (hypotenuse).
sin^2(theta) = (CD/BD)^2 = (CD/BC)^2
Substitute CD^2 = 108 and BC = 12:
sin^2(theta) = (sqrt(108)/12)^2
Simplifying further:
sin^2(theta) = (sqrt(108))^2 / (12)^2
sin^2(theta) = 108 / 144
sin^2(theta) = 3/4
Similarly, cos^2(theta) = 1 - sin^2(theta)
cos^2(theta) = 1 - (3/4)
cos^2(theta) = 1/4
5. Finally, substitute the values of sin^2(theta) and cos^2(theta) in the expression:
sin^2(theta) + cos^2(theta) - 1 = (3/4) + (1/4) - 1
= 3/4 + 1/4 - 4/4
= 4/4 - 4/4
= 0
Therefore, the value of sin^2(theta) + cos^2(theta) - 1 is 0.