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3 years ago

Given: △ABC, m∠A=60° m∠C=45°, AB=8 Find: Perimeter of △ABC, Area of △ABC

Mathematics

2 answers:

jarptica [38.1K]3 years ago

6 0

We are given

△ABC, m∠A=60° m∠C=45°, AB=8

Firstly, we will find all angles and sides

Calculation of angle B:

we know that sum of all angles is 180

m∠A+ m∠B+m∠C=180

we can plug values

60°+ m∠B+45°=180

m∠B=75°

Calculation of BC:

we can use law of sines

$\frac{AB}{sin(C)}=\frac{BC}{sin(A)}$

now, we can plug values

$\frac{8}{sin(45)}=\frac{BC}{sin(60)}$

$BC=\frac{8}{sin(45)} \times sin(60)$

$BC=9.798$

Calculation of AC:

$\frac{AB}{sin(C)}=\frac{AC}{sin(B)}$

now, we can plug values

$\frac{8}{sin(45)}=\frac{AC}{sin(75)}$

$AC=\frac{8}{sin(45)} \times sin(75)$

$AC=10.928$

Perimeter:

$p=AB+BC+AC$

we can plug values

$p=10.928+8+9.798$

$p=28.726$

Area:

we can use formula

$A=\frac{1}{2}AB \times AC \times sin(A)$

now, we can plug values

$A=\frac{1}{2}8 \times 10.928 \times sin(60)$

$A=37.85570$ ...............Answer

igomit [66]3 years ago

6 0

The perimeter of triangle of ABC is $\boxed{28.73}.$

Further explanation:

Given:

The measure of angle A is $\angle A = {60^ \circ }.$

The measure of angle C is $\angle C = {45^ \circ }.$

The length of side AB is $AB = 8$

Calculation:

The sum of all angles of a triangle is ${180^ \circ }.$

$\begin{aligned}\angle A + \angle B + \angle C&={180^ \circ }\\{60^ \circ } + \angle B + {45^ \circ }&= {180^ \circ }\\{105^ \circ }+\angle B&= {180^ \circ }\\\angleB&= {180^ \circ } - {105^ \circ }\\\angleB&= {75^ \circ }\\\end{aligned}$

The sine rule in triangle ABC can be expressed as,

$\begin{aligned}\frac{{BC}}{{\sin {{60}^ \circ }}}&=\frac{8}{{\sin {{45}^ \circ }}}\\BC&= \frac{8}{{\frac{1}{{\sqrt2 }}}} \times \frac{{\sqrt 3 }}{2}\\BC &= 9.80\\\end{aligned}$

The length of AC can be calculated as follows,

$\begin{aligned}\frac{{AB}}{{\sin {{45}^ \circ }}}&=\frac{{AC}}{{\sin {{75}^ \circ }}}\\\frac{8}{{\sin {{45}^ \circ }}}\times \sin {75^ \circ }&= AC\\10.93& = AC\\\end{aligned}$

The perimeter of triangle ABC can be obtained as follows,

$\begin{aligned}{\text{Perimeter}}&= AB + BC + AC\\&= 8 + 9.80 + 10.93\\&= 28.73\\\end{aligned}$

The area of triangle ABC can be obtained as follows,

$\begin{aligned}{\text{Area}}&=\frac{1}{2} \times AB \times AC \times \sin \left( A \right)\\&= \frac{1}{2}\times 8 \times 10.93 \times \sin {60^ \circ }\\&= 4\times 10.93 \times \frac{{\sqrt3 }}{2}\\&= 37.86\\\end{aligned}$