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The angles of a triangle are in the rati...

The angles of a triangle are in the ratio 3 : 5 : 10, the ratio of the smallest side to the greatest side is

A

`1 : sin10^(@)`

B

`1 : 2 sin 10^(@)`

C

`1 : cos 10^(@)`

D

`1 : 2 cos10^(@)`

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The correct Answer is:
To solve the problem of finding the ratio of the smallest side to the greatest side of a triangle whose angles are in the ratio 3:5:10, we can follow these steps: ### Step 1: Define the Angles Let the angles of the triangle be represented as: - \( A = 3x \) - \( B = 5x \) - \( C = 10x \) ### Step 2: Set Up the Equation Since the sum of the angles in a triangle is always \( 180^\circ \), we can set up the equation: \[ 3x + 5x + 10x = 180 \] ### Step 3: Solve for \( x \) Combine like terms: \[ 18x = 180 \] Now, divide both sides by 18: \[ x = 10 \] ### Step 4: Calculate the Angles Now substitute \( x \) back into the expressions for the angles: - \( A = 3x = 3 \times 10 = 30^\circ \) - \( B = 5x = 5 \times 10 = 50^\circ \) - \( C = 10x = 10 \times 10 = 100^\circ \) ### Step 5: Identify the Smallest and Greatest Angles From the calculated angles: - The smallest angle is \( A = 30^\circ \). - The greatest angle is \( C = 100^\circ \). ### Step 6: Use the Law of Sines to Find the Ratio of Sides According to the Law of Sines: \[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \] Where \( a \), \( b \), and \( c \) are the sides opposite to angles \( A \), \( B \), and \( C \) respectively. ### Step 7: Express the Ratio of the Smallest Side to the Greatest Side We want to find the ratio \( \frac{a}{c} \): \[ \frac{a}{c} = \frac{\sin A}{\sin C} \] ### Step 8: Substitute the Values of the Angles Substituting the values: \[ \frac{a}{c} = \frac{\sin 30^\circ}{\sin 100^\circ} \] ### Step 9: Calculate the Sine Values Using known sine values: - \( \sin 30^\circ = \frac{1}{2} \) - \( \sin 100^\circ = \sin(90^\circ + 10^\circ) = \cos 10^\circ \) ### Step 10: Write the Final Ratio Thus, the ratio becomes: \[ \frac{a}{c} = \frac{\frac{1}{2}}{\cos 10^\circ} = \frac{1}{2 \cos 10^\circ} \] ### Conclusion The ratio of the smallest side to the greatest side of the triangle is: \[ \frac{1}{2 \cos 10^\circ} \]

To solve the problem of finding the ratio of the smallest side to the greatest side of a triangle whose angles are in the ratio 3:5:10, we can follow these steps: ### Step 1: Define the Angles Let the angles of the triangle be represented as: - \( A = 3x \) - \( B = 5x \) - \( C = 10x \) ...
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OBJECTIVE RD SHARMA-PROPERTIES OF TRIANGLES AND CIRCLES CONNECTED WITH THEM-Chapter Test
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  2. If the sides of a triangle are in the ratio 3 : 7 : 8, then find R : r

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  3. The area of the reactangle polygen of n sides is (where R is the radiu...

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  4. If the angles of a rectangle are 30^(@) and 45^(@) and the included si...

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  5. In a triagnle ABC, angle B=pi/3 " and " angle C = pi/4 let D divide ...

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  6. If A is the area and 2s the sum of the sides of a triangle,then

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  7. If in a triangle ABC, right angled at B, s-a=3, s-c=2, then the values...

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  8. If the sides of a triangle are a, b and sqrt(a^(2) + ab + b^(2)), then...

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  9. In a DeltaA B Csum(b+c)tanA/2tan((B-C)/2)=

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  10. In triangle ABC, angleA=pi/3 and b:c =2:3, tan theta=sqrt3/5, 0 lt the...

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  11. In a DeltaABC, AD is the altitude from A. Given bgtc ,angleC=23^(@) an...

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  12. If the angles A, B, C (in that order) of triangle ABC are in arithmeti...

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  13. If the radius of the incircle of a triangle withits sides 5k, 6k and 5...

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  14. Two sides of a triangle are 2sqrt2 and 2sqrt3cm and the angle opposite...

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  15. In a triangleABC, a=13cm, b=12 and c=5cm The distance of A from BC is

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  16. In a triangleABC,B=pi/8, C=(5pi)/(8). The altitude from A to the side ...

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  17. In a DeltaABC, A=(2pi)/3, b-c=3sqrt3 cm and area(DeltaABC)=(9sqrt3)/2 ...

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  18. In DeltaABC if a=(b-c)sectheta then (2sqrt(bc))/(b-c)sin(A/2)=

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  19. In a DeltaABC, (a + b + c) (b + c - a) = lambda bc. (where symbols ha...

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  20. If in DeltaABC, a=2b and A=3B, then A is equal to

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  21. Let the angles A , B and C of triangle A B C be in AdotPdot and let b ...

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