Question

if 17 cos A = 15 then find the value
of tan A +2 secA​

Answer 1

Answer:

17cosA=15

cosA=15/17

cosA=B/H,B=15,H=17,P=?

AB=17 ,AC= ? BC=15

in triangle ABC (apply PGT)

ABsquare=ACsquare+BCsquare

(17)^2=AC^2+15^2

289=AC^2+225

AC^2=289-225

AC^2=64

AC=√64

AC=8

AC=P=8

tanA=P/B

tanA=8/15

secA=H/B

secA=17/15

tanA+2secA

=8/15+2(17/15)

=8/15+34/15

=8+34/15

=42/15

=14/5

Answer 2

$\huge\bold{\blue{Q}\green{U}\pink{E}\orange{S}\purple{T}\red{I}\green{O}\pink{N}}$

☞If 17 cos A = 15 then, find the value of tan A +2 secA.

$\huge\bold{\blue{A}\green{N}\pink{S}\orange{W}\purple{E}\red{R}}$

➜GIVEN THAT:

17COSA=15

$</strong><strong>{</strong><strong>{</strong><strong>cos(</strong><strong>a</strong><strong>)</strong><strong> </strong><strong>= \dfrac{15}{17} \\</strong><strong> </strong><strong>\</strong><strong>\</strong><strong> \\ = > </strong><strong>base</strong><strong> = 15 \: </strong><strong>\</strong><strong>\</strong><strong> </strong><strong>\: </strong><strong>hypotenuse</strong><strong> = 17 \: \: </strong><strong>\</strong><strong>\</strong><strong> </strong><strong>\\ </strong><strong>\</strong><strong>\</strong><strong> </strong><strong>perpendicular</strong><strong> = \sqrt{ {h}^{2} - {b}^{2} } </strong><strong>\</strong><strong>\</strong><strong> </strong><strong>= \sqrt{ {17}^{2} - {15}^{2} } \\ = \sqrt{64} </strong><strong>\</strong><strong>\</strong><strong> </strong><strong>= 8 \\</strong><strong> </strong><strong>\</strong><strong>\</strong><strong> \\ \\ tan(</strong><strong>A</strong><strong>) = \</strong><strong>d</strong><strong>frac{p}{h} = \</strong><strong>d</strong><strong>frac{8}{1</strong><strong>5</strong><strong>} \\ \\ \</strong><strong>sec(a)</strong><strong> </strong><strong> = \</strong><strong>d</strong><strong>frac{h}{b} = \</strong><strong>d</strong><strong>frac{17}{15 } \\ \\ \\ = > \</strong><strong>tan(a)</strong><strong> + 2 \</strong><strong>sec(a)</strong><strong> = \</strong><strong>d</strong><strong>frac{8}{1</strong><strong>5</strong><strong>} + \</strong><strong>d</strong><strong>frac{2 \times 17}{15} </strong><strong>\</strong><strong>\</strong><strong> </strong><strong>\\ = \</strong><strong>d</strong><strong>frac{8}{1</strong><strong>5</strong><strong>} + \</strong><strong>d</strong><strong>frac{34}{15}</strong><strong>\</strong><strong>\</strong><strong> \\ = \</strong><strong>d</strong><strong>frac{</strong><strong>8</strong><strong>+</strong><strong>1</strong><strong>7</strong><strong>}{</strong><strong>1</strong><strong>5</strong><strong>} </strong><strong>\</strong><strong>\</strong><strong> \\ = \</strong><strong>d</strong><strong>frac{</strong><strong>2</strong><strong>5</strong><strong>}{</strong><strong>1</strong><strong>5</strong><strong>}</strong><strong>\</strong><strong>\</strong><strong> </strong><strong>\</strong><strong>\</strong><strong> </strong><strong>=</strong><strong> </strong><strong> </strong><strong>\</strong><strong>d</strong><strong>f</strong><strong>r</strong><strong>a</strong><strong>c</strong><strong>{</strong><strong>5</strong><strong>}</strong><strong>{</strong><strong>3</strong><strong>}</strong><strong>}</strong><strong>}</strong><strong>$

HOPE IT HELPS