Question

if 5Cos A + 12sinA = 13 find the value of tanA​

Answer 1

Answer:

The answer will be -8.

Step-by-step explanation:

Given;

5cos A + 12sinA = 13

Add 7cos A on the both side of the given equation;

7cos A + 5cos A + 12sinA = 13 + 7cos A

12cos A + 12sin A = 13 + 7cos A

12(cos A + sin A) = 13 + 7cos A

12 (1) = 13 + 7cos A. (since, cos A + sin A = 1)

12 - 13 = 7cos A

-1/7 = cos A

cos A = -1/7

Now,minus 7sin A on the both side of the given equation;

5cos A + 12sinA - 7sin A= 13 - 7sin A

5cos A + 5sin A = 13 - 7sin A

5(cos A + sin A) = 13 - 7sin A

5 (1) = 13 - 7sin A. (since, cos A + sin A = 1)

7sin A = 13 - 5

sin A = 8/7

Now, tan A = sin A/cos A

Thus, tan A = (8/7) / (-1/7)

= 8/-1

= -8.

That's all.

Answer 2

$\large\underline{\sf{Given- }}$

• 5 cosA + 12 sinA = 13

$\large\underline{\sf{To\:Find - }}$

• tanA

$\begin{gathered}\Large{\sf{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\boxed{\bf{\: \dfrac{1}{cosA} = secA}}$

$\boxed{ \bf{ \: \dfrac{1}{sinA} = cosecA}}$

$\boxed{ \bf{ \: \dfrac{sinA}{cosA} = tanA}}$

$\boxed{ \bf{ \: {sec}^{2}A - {tan}^{2}A = 1}}$

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:5cosA + 12sinA = 13$

• On dividing both sides by cosA, we get

$\rm :\longmapsto\:5\dfrac{cosA}{cosA} + 12\dfrac{sinA}{cosA} = \dfrac{13}{cosA}$

$\rm :\longmapsto\:5 + 12tanA = 13secA$

• On squaring both sides, we get

$\rm :\longmapsto\: {(5 + 12tanA)}^{2} = {(13secA)}^{2}$

$\rm :\longmapsto\:25 + 144 {tan}^{2}A + 120tanA = 169 {sec}^{2}A$

$\rm :\longmapsto\:25 + 144 {tan}^{2}A + 120tanA = 169 (1 + {tan}^{2}A)$

$\rm :\longmapsto\:25 + 144 {tan}^{2}A + 120tanA = 169 + 169 {tan}^{2}A$

$\rm :\longmapsto\:25 {tan}^{2}A - 120tanA + 144 = 0$

$\rm :\longmapsto\: {(5tanA)}^{2} + {(12)}^{2} - 2 \times 5tanA \times 12 = 0$

$\rm :\longmapsto\: {(5tanA - 12)}^{2} = 0$

$\: \: \: \: \: \: \: \: \: \: \boxed{ \bf{ \: \because \: {x}^{2} + {y}^{2} - 2xy = {(x - y)}^{2}}}$

$\rm :\longmapsto\:5tanA - 12 = 0$

$\rm :\longmapsto\:5tanA = 12$

$\bf\implies \:tanA = \dfrac{12}{5}$

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Additional Information :-

$\boxed{ \bf{ \: {sin}^{2}x + {cos}^{2}x = 1}}$

$\boxed{ \bf{ \: {sin}^{2}x = 1 - {cos}^{2}x}}$

$\boxed{ \bf{ \: {cos}^{2} x = 1 - {sin}^{2} x}}$

$\boxed{ \bf{ \: {cosec}^{2} x - {cot}^{2} x = 1}}$

$\boxed{ \bf{ \: {cosec}^{2} x = 1 + {cot}^{2} x}}$

$\boxed{ \bf{ \: {cot}^{2} x = {cosec}^{2} x - 1}}$