Question

ABC right angled at B AB = 5cm and ACB = 30 Determine the lengths of the side BC and AC​

Answer 1

Answer:-

$\large\sf{\frac{AB}{BC}=tanC}$

$\large\sf{i.e.,\:\frac{5}{BC}=tan30°=\frac{1}{√3}}$

$\large\sf{which\:gives\:5√3cm}$

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$\small\sf{to\:find\:the\:length\:of\:the\:side\:AC,}$

$\large\sf{we\:consider,}$

$\longrightarrow$$\large\sf{sin30°=\frac{AB}{AC}}$

$\longrightarrow$$\large\sf{\frac{1}{2}=\frac{5}{AC}}$

$\longrightarrow$$\large\sf{AC=10cm}$

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$\large\sf{By\:using\:Pythagoras\:theorem,}$

$\longrightarrow$$\large\sf{AC=√{AB}^{2}+{BC}^{2}}$

$\longrightarrow$$\large\sf{√{5}^{2}+{(5√3)}^{2}}$

$\longrightarrow$$\large\sf{10cm}$

Answer 2

Answer:-

\large\sf{\frac{AB}{BC}=tanC}

BC

AB

=tanC

\large\sf{i.e.,\:\frac{5}{BC}=tan30°=\frac{1}{√3}}i.e.,

BC

5

=tan30°=

√3

1

\large\sf{which\:gives\:5√3cm}whichgives5√3cm

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\small\sf{to\:find\:the\:length\:of\:the\:side\:AC,}tofindthelengthofthesideAC,

\large\sf{we\:consider,}weconsider,

\longrightarrow⟶ \large\sf{sin30°=\frac{AB}{AC}}sin30°=

AC

AB

\longrightarrow⟶ \large\sf{\frac{1}{2}=\frac{5}{AC}}

2

1

=

AC

5

\longrightarrow⟶ \large\sf{AC=10cm}AC=10cm

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\large\sf{By\:using\:Pythagoras\:theorem,}ByusingPythagorastheorem,

\longrightarrow⟶ \large\sf{AC=√{AB}^{2}+{BC}^{2}}AC=√AB

2

+BC

2

\longrightarrow⟶ \large\sf{√{5}^{2}+{(5√3)}^{2}}√5

2

+(5√3)

2

\longrightarrow⟶ \large\sf{10cm}10cm