Question

please answer these question
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Answer 1

Given Question :-

In triangle ABC, AB = 7 cm, BC = 12 cm and angle B = 40°. Find

1. Area of triangle ABC

2. Length of AC.

Given that sin40° = 0.6428 and cos40° = 0.7660

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

Let assume that the sides of triangle ABC is represented as

~ AB = c cm

~ BC = a cm

~ CA = b cm

Now, given that,

~ AB = c = 7 cm

~ BC = a = 12 cm

Now, we know that,

Area of triangle ABC is given by

$\rm \: = \:\dfrac{1}{2}ac \: sinB$

$\rm \: = \:\dfrac{1}{2} \times 7 \times 12 \times \: sin40 \degree$

$\rm \: = \:7 \times 6 \times 0.6428$

$\rm \: = \:26.99 \: \approx \: 27 \: {cm}^{2}$

Now, To calculate the length of AC

We know,

~ By cosine Law

$\boxed{ \bf{ \: cosB = \frac{ {a}^{2} + {c}^{2} - {b}^{2} }{2ac}}}$

So, on substituting the values, we get

$\rm :\longmapsto\:cos40 \degree \: = \: \dfrac{ {7}^{2} + {12}^{2} - {b}^{2} }{2 \times 7 \times 12}$

$\rm :\longmapsto\:0.766 = \dfrac{49 + 144 - {b}^{2} }{168}$

$\rm :\longmapsto\:128.688 = 193 - {b}^{2}$

$\rm :\longmapsto\:128.688 - 193 = - {b}^{2}$

$\rm :\longmapsto\: - {b}^{2} = - 64.312$

$\rm :\longmapsto\: {b}^{2} = 64.312$

$\bf\implies \:b = 8.038 \: \approx \: 8.04 \: cm$

Additional Information :-

$\boxed{ \bf{ \: \frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC}}}$

$\boxed{ \bf{ \: cosA = \frac{ {b}^{2} + {c}^{2} - {a}^{2} }{2bc}}}$

$\boxed{ \bf{ \: cosC = \frac{ {b}^{2} + {a}^{2} - {c}^{2} }{2ab}}}$

$\boxed{ \bf{ \: a = b \: cosC + c \: cosB}}$

$\boxed{ \bf{ \: b = a \: cosC + c \: cosA}}$

$\boxed{ \bf{ \: c = a \: cosB + b \: cosA}}$