Question

Calculate the size of angel X you must show all you working out!

Answer 1

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

Points B, C, D lie on a straight line.

AC is perpendicular to BD.

BC = 6 cm, BD = 22 cm

Area of triangle ABC is 42 square cm.

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

The value of angle x

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

Given that,

• AC is perpendicular to BD

• BC = 6 cm

• Area of triangle ABC = 42 square cm.

We know

$\red{\rm :\longmapsto\:Area_{\triangle ABC} = \dfrac{1}{2} \times BC \times AC}$

$\red{\rm :\longmapsto\:42 = \dfrac{1}{2} \times 6 \times AC}$

$\red{\rm :\longmapsto\:42 = 3 \times AC}$

$\red{\rm :\longmapsto\: AC = 14 \: cm}$

Now, Consider

$\rm :\longmapsto\:BD = BC + AD$

$\rm :\longmapsto\:22 = 6 + AD$

$\rm :\longmapsto\:22 - 6 = AD$

$\rm :\longmapsto\:AD = 16 \: cm$

$\rm :\longmapsto\:In \: right \: \triangle \: ADC$

Using Trigonometric ratios, we have

$\rm :\longmapsto\:tanx = \dfrac{AC}{AD}$

$\rm :\longmapsto\:tanx = \dfrac{14}{16}$

$\rm :\longmapsto\:tanx = \dfrac{7}{8}$

$\rm :\longmapsto\:tanx = 0.875$

$\rm :\longmapsto\:x = 41 \degree \: 12' \: approx$

$\bf\implies \:x = 41.20 \degree \: approx.$

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Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1