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# ABCD is a cyclic quadrilateral such that 12 tan(A) – 5 = 0 and 5 cos(B) + 3 =0. Find the value of 39(cos(C)+tan(D)).(a). -16(b). 16(c). -14(d). 14  Answer Verified
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.

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