Question

triangle ABC is right angled at B and tan a = 4/3 if AC = 15 CM the length of Ab is​

Answer 1

Answer:

BA=3√15

Step-by-step explanation:

tan a=BC/BA

4/3 =BC/BA

4BA/3=BC

Let BA=@

w.k.t, AC^2=BA^2+BC^2

225=@^2+(4@/3)^2

225=@^2+16/9×@^2

225=(25@^2)÷9

225×9/15=@^2

145=@^2

@=√145

@=√(9×15)

@=3√15

Therefore BA=3√15

Answer 2

Concept: The relationship between triangle side lengths and angles is examined in the math discipline of trigonometry. Six different trigonometric functions can be applied to a common angle. They are known by the designations sine (sin), cosine (cos), tangent (tan), cotangent (cot), secant (sec), and cosecant (cosec).

Given: a triangle ABC right angled at B

           $tan A = \frac{4}{3}$

           AC = 15cm

To find: length of AB

Solution:

Refer the figure below for clear explanation

Let us suppose the length of AB as x cm

We know $tan A = \frac{perpendicular}{base}$

               $tan A = \frac{4}{3}$  

perpendicular to A is BC and base of A is AB

therefore, the ratio between the perpendicular BC and the base AB is 4k:3k [ k being any constant]

Applying the Pythagoras theorem,

$(perpendicular)^2 + (base)^2 = (hypotenuse)^2\\(4k)^2 + (3k)^2 = (hypotenuse)^2\\16k^2 + 9k^2 = (hypotenuse)^2\\25k^2 = (hypotenuse)^2\\5k = hypotenuse$

From the figure, 5k = 15      [ in the figure, hypotenuse = 15]

$k = \frac{15}{5}$

k = 3

Now, substituting the value of k in the ratio 4k:3k, we get

BC = 4k = 4 × 3 = 12

AB = 3k = 3 × 3 = 9

Hence, the length of AB = 9 cm.

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