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Given that Delta = 6, r(1) = 2,r(2)=3, ...

Given that `Delta = 6, r_(1) = 2,r_(2)=3, r_(3) = 6` Difference between the greatest and the least angles is

A

`cos^(-1).(4)/(5)`

B

`tan^(-1).(3)/(4)`

C

`cos^(-1).(3)/(5)`

D

none of these

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To solve the problem, we need to find the difference between the greatest and the least angles of a triangle given the values of the area (Δ) and the inradii (r1, r2, r3). ### Step-by-Step Solution: 1. **Identify the Given Values**: - Area of the triangle (Δ) = 6 - Inradius values: r1 = 2, r2 = 3, r3 = 6 2. **Use the Formula for Inradius**: The inradius \( r \) of a triangle can be expressed in terms of the area \( Δ \) and the semi-perimeter \( s \): \[ r = \frac{Δ}{s} \] For each inradius, we can write: - \( r_1 = \frac{Δ}{s - a} \) - \( r_2 = \frac{Δ}{s - b} \) - \( r_3 = \frac{Δ}{s - c} \) 3. **Set Up the Equations**: Plugging in the values, we have: - \( 2 = \frac{6}{s - a} \) → \( s - a = 3 \) → \( s = a + 3 \) (Equation 1) - \( 3 = \frac{6}{s - b} \) → \( s - b = 2 \) → \( s = b + 2 \) (Equation 2) - \( 6 = \frac{6}{s - c} \) → \( s - c = 1 \) → \( s = c + 1 \) (Equation 3) 4. **Equate the Semi-Perimeter**: From the three equations, we can equate the expressions for \( s \): \[ a + 3 = b + 2 = c + 1 \] 5. **Express \( a, b, c \) in Terms of Each Other**: Rearranging gives: - From Equation 1: \( a = s - 3 \) - From Equation 2: \( b = s - 2 \) - From Equation 3: \( c = s - 1 \) 6. **Substitute and Solve**: Using \( s = a + 3 \): - Substitute \( b \) and \( c \): - \( b = (a + 3) - 2 = a + 1 \) - \( c = (a + 3) - 1 = a + 2 \) 7. **Find the Angles**: Since we have \( a, b, c \) in terms of \( a \): - \( a = 3 \) - \( b = 4 \) - \( c = 5 \) 8. **Determine the Greatest and Least Angles**: In a triangle, the angle opposite the longest side is the greatest angle. Here, side lengths are: - \( a = 3 \) - \( b = 4 \) - \( c = 5 \) (hypotenuse) Thus, the greatest angle \( C \) is \( 90^\circ \) (as it is a right triangle), and the smallest angle \( A \) can be calculated using the sine rule or cosine rule. 9. **Calculate the Smallest Angle**: Using the sine function: \[ \sin A = \frac{a}{c} = \frac{3}{5} \] Therefore, \( A = \sin^{-1}\left(\frac{3}{5}\right) \). 10. **Find the Difference**: The difference between the greatest and the least angle is: \[ 90^\circ - A = 90^\circ - \sin^{-1}\left(\frac{3}{5}\right) \] ### Final Answer: The difference between the greatest and the least angles is: \[ 90^\circ - \sin^{-1}\left(\frac{3}{5}\right) \]

To solve the problem, we need to find the difference between the greatest and the least angles of a triangle given the values of the area (Δ) and the inradii (r1, r2, r3). ### Step-by-Step Solution: 1. **Identify the Given Values**: - Area of the triangle (Δ) = 6 - Inradius values: r1 = 2, r2 = 3, r3 = 6 ...
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CENGAGE ENGLISH-PROPERTIES AND SOLUTIONS OF TRIANGLE-Linked comprehension type
  1. Given that Delta = 6, r(1) = 3, r(3) = 6 Circumradius R is equal to

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  2. Given that Delta = 6, r(1) = 3, r(3) = 6 Inradius is equal to

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  3. Given that Delta = 6, r(1) = 2,r(2)=3, r(3) = 6 Difference between t...

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  4. Let a = 6, b = 3 and cos (A -B) = (4)/(5) Area (in sq. units) of the...

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  5. Let a = 6, b = 3 and cos (A -B) = (4)/(5) Angle C is equal to

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  6. Let a = 6, b = 3 and cos (A -B) = (4)/(5) Value of sin A is equal to

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  8. Let ABC be an acute angled triangle with orthocenter H.D, E, and F are...

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  9. Let ABC be an acute angled triangle with orthocenter H.D, E, and F are...

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  10. Let O be a point inside a triangle A B C such that /O A B=/O B C=/O C ...

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  11. find the principle value of cos^(-1)((sqrt3)/2)

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  12. Let O be a point inside DeltaABC such that angleAOB = angleBOC = ang...

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  15. Given an isoceles triangle with equal side of length b and angle alpha...

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  16. Incircle of DeltaABC touches the sides BC, AC and AB at D, E and F, re...

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  17. Incircle of DeltaABC touches the sides BC, AC and AB at D, E and F, re...

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  18. Incircle of DeltaABC touches the sides BC, AC and AB at D, E and F, re...

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  19. Bisectors of angles A, B and C of a triangle ABC intersect its circum...

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