Question

In triangle abc, right angled at B, AC-BC=8cm and AB=12cm. Find the value of 4cot A+sec C

Answer 1

Answer:

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Answer 2

The value of 4cot A+sec C is 61/5.

Given:

In triangle ABC right angled at B

AC-BC=8cm

AB=12cm

To Find:

The value of 4cot A+sec C.

Solution:

From Pythagoras' theorem, we know,

AC² = AB² + BC² ---------(1)

AC - hypotenuse side

BC - adjacent side

AB - opposite side

where AC = BC + 8 substitute in (1)

(BC + 8)² = AB² + BC²

BC² + 16BC +64 = AB² + BC²

16BC = 144 - 64

16BC = 80

BC = 5 cm substitute in AC = BC + 8

AC = 5 + 8

      = 13 cm

Cot A = adjacent side / opposite side from point A

          = AB/BC

          = 12/5

Sec C = hypotenuse / adjacent side from point C

          = AC/BC

          = 13/5

4cot A+sec C = $4*\frac{12}{5} + \frac{13}{5}$

                       = $\frac{48}{5} + \frac{13}{5}$

                       = $\frac{61}{5}$

Therefore, The value of 4cot A+sec C is 61/5.

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