Answer
Verified
415.8k+ views
Hint: The sum of opposite angles of the cyclic quadrilateral is 180 degrees. Use this to find cos(C) and tan(D). Then, we can find the value of the required expression.
Complete step-by-step answer:
Given that ABCD is a cyclic quadrilateral. We know that the opposite angles of a cyclic quadrilateral add up to 180 degrees. The pair of opposite angles of the cyclic quadrilateral ABCD are A and C and B and D. Hence, the relation between them is as follows:
\[A + C = 180\]
\[C = 180 - A.........(1)\]
\[B + D = 180\]
\[D = 180 - B.........(2)\]
We can substitute equation (1) and equation (2) in the given expressions.
\[12\tan A - 5 = 0\]
\[\tan A = \dfrac{5}{{12}}\]
\[\tan (180 - C) = \dfrac{5}{{12}}\]
We know that the value of \[\tan (180 - x)\] is -tan(x).
\[\tan C = - \dfrac{5}{{12}}...........(3)\]
The second expression can be written as:
\[5\cos B + 3 = 0\]
\[\cos B = - \dfrac{3}{5}\]
\[\cos (180 - D) = - \dfrac{3}{5}\]
We know that \[\cos (180 - x)\] is -cos(x), then, we have:
\[\cos D = \dfrac{3}{5}..............(4)\]
From equation (3), let us find out cos(C).
We know that \[{\sec ^2}x = 1 + {\tan ^2}x\]. Using this in the above equation, we get:
\[{\sec ^2}C = 1 + {\tan ^2}C\]
\[{\sec ^2}C = 1 + {\left( { - \dfrac{5}{{12}}} \right)^2}\]
Simplifying the square, we get:
\[{\sec ^2}C = 1 + \dfrac{{25}}{{144}}\]
\[{\sec ^2}C = \dfrac{{144 + 25}}{{144}}\]
\[{\sec ^2}C = \dfrac{{169}}{{144}}\]
We know that 169 is square of 13 and 144 is square of 12.
\[\sec C = \pm \dfrac{{13}}{{12}}\]
Angle C lies in the second quadrant since tan(C) is negative. Hence, sec(C) is also negative.
\[\sec C = - \dfrac{{13}}{{12}}\]
We know that \[\cos x = \dfrac{1}{{\sec x}}\], hence we have:
\[\cos C = - \dfrac{{12}}{{13}}............(5)\]
From equation (4), let us find out tan(D).
We know that, \[\sec x = \dfrac{1}{{\cos x}}\], hence, we have:
\[\sec D = \dfrac{5}{3}\]
We know that \[{\tan ^2}x = {\sec ^2}x - 1\]. Using this in the above equation, we get:
\[{\tan ^2}D = {\left( {\dfrac{5}{3}} \right)^2} - 1\]
\[{\tan ^2}D = \dfrac{{25}}{9} - 1\]
\[{\tan ^2}D = \dfrac{{25 - 9}}{9}\]
\[{\tan ^2}D = \dfrac{{16}}{9}\]
We know that 16 is square of 4 and 9 is square of 3, hence we have:
\[\tan D = \pm \dfrac{4}{3}\]
Angle D lies in the first quadrant since cos(D) is positive, hence, tan(D) is also positive.
\[\tan D = \dfrac{4}{3}...........(6)\]
We have to find the value of \[39(\cos C + \tan D)\], using equation (5) and equation (6), we get:
\[39(\cos C + \tan D) = 39\left( { - \dfrac{{12}}{{13}} + \dfrac{4}{3}} \right)\]
\[39(\cos C + \tan D) = 39\left( {\dfrac{{ - 36 + 52}}{{39}}} \right)\]
\[39(\cos C + \tan D) = 16\]
Hence, the correct answer is option (b).
Note: If you do not write the plus and minus sign correctly when taking square roots, you might get option (a) as the correct answer, which is wrong. Use the quadrants to check if the square root value is positive or negative.
Complete step-by-step answer:
Given that ABCD is a cyclic quadrilateral. We know that the opposite angles of a cyclic quadrilateral add up to 180 degrees. The pair of opposite angles of the cyclic quadrilateral ABCD are A and C and B and D. Hence, the relation between them is as follows:
\[A + C = 180\]
\[C = 180 - A.........(1)\]
\[B + D = 180\]
\[D = 180 - B.........(2)\]
We can substitute equation (1) and equation (2) in the given expressions.
\[12\tan A - 5 = 0\]
\[\tan A = \dfrac{5}{{12}}\]
\[\tan (180 - C) = \dfrac{5}{{12}}\]
We know that the value of \[\tan (180 - x)\] is -tan(x).
\[\tan C = - \dfrac{5}{{12}}...........(3)\]
The second expression can be written as:
\[5\cos B + 3 = 0\]
\[\cos B = - \dfrac{3}{5}\]
\[\cos (180 - D) = - \dfrac{3}{5}\]
We know that \[\cos (180 - x)\] is -cos(x), then, we have:
\[\cos D = \dfrac{3}{5}..............(4)\]
From equation (3), let us find out cos(C).
We know that \[{\sec ^2}x = 1 + {\tan ^2}x\]. Using this in the above equation, we get:
\[{\sec ^2}C = 1 + {\tan ^2}C\]
\[{\sec ^2}C = 1 + {\left( { - \dfrac{5}{{12}}} \right)^2}\]
Simplifying the square, we get:
\[{\sec ^2}C = 1 + \dfrac{{25}}{{144}}\]
\[{\sec ^2}C = \dfrac{{144 + 25}}{{144}}\]
\[{\sec ^2}C = \dfrac{{169}}{{144}}\]
We know that 169 is square of 13 and 144 is square of 12.
\[\sec C = \pm \dfrac{{13}}{{12}}\]
Angle C lies in the second quadrant since tan(C) is negative. Hence, sec(C) is also negative.
\[\sec C = - \dfrac{{13}}{{12}}\]
We know that \[\cos x = \dfrac{1}{{\sec x}}\], hence we have:
\[\cos C = - \dfrac{{12}}{{13}}............(5)\]
From equation (4), let us find out tan(D).
We know that, \[\sec x = \dfrac{1}{{\cos x}}\], hence, we have:
\[\sec D = \dfrac{5}{3}\]
We know that \[{\tan ^2}x = {\sec ^2}x - 1\]. Using this in the above equation, we get:
\[{\tan ^2}D = {\left( {\dfrac{5}{3}} \right)^2} - 1\]
\[{\tan ^2}D = \dfrac{{25}}{9} - 1\]
\[{\tan ^2}D = \dfrac{{25 - 9}}{9}\]
\[{\tan ^2}D = \dfrac{{16}}{9}\]
We know that 16 is square of 4 and 9 is square of 3, hence we have:
\[\tan D = \pm \dfrac{4}{3}\]
Angle D lies in the first quadrant since cos(D) is positive, hence, tan(D) is also positive.
\[\tan D = \dfrac{4}{3}...........(6)\]
We have to find the value of \[39(\cos C + \tan D)\], using equation (5) and equation (6), we get:
\[39(\cos C + \tan D) = 39\left( { - \dfrac{{12}}{{13}} + \dfrac{4}{3}} \right)\]
\[39(\cos C + \tan D) = 39\left( {\dfrac{{ - 36 + 52}}{{39}}} \right)\]
\[39(\cos C + \tan D) = 16\]
Hence, the correct answer is option (b).
Note: If you do not write the plus and minus sign correctly when taking square roots, you might get option (a) as the correct answer, which is wrong. Use the quadrants to check if the square root value is positive or negative.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Select the correct plural noun from the given singular class 10 english CBSE
The sum of three consecutive multiples of 11 is 363 class 7 maths CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
How many squares are there in a chess board A 1296 class 11 maths CBSE
How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
There are n locks and n matching keys If all the locks class 11 maths CBSE
I buy a TV for Rs 10000 and sell it at a profit of-class-7-maths-CBSE