In figure- AB - 9- BC - 5- and - C - 50- Calculate the area of triangle ABC- to the nearest tenth-

The correct option is A 21.7 By sine rule, we get #160; AB sin 50 = BC sin A #160; ⇒sin A = #160; 5 9 ×sin50=0.426 ⇒ A = 25.2 degre ...

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Standard XII

Mathematics

Integration of Piecewise Continuous Functions

In figure, ...

Question

In figure, $A B = 9, B C = 5$ , and $∠ C = 50^{\circ}$ . Calculate the area of triangle $A B C$ , to the nearest tenth.

21.7

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22.5

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36.4

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40.2

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Solution

The correct option is A $21.7$
By sine rule, we get
$\frac{A B}{sin 50} = \frac{B C}{sin A}$
$\Rightarrow sin A = \frac{5}{9} \times sin 50 = 0.426$
$\Rightarrow A = 25.2$ degrees
Therefore $B = 180 - 50 - 25.2 = 104.8$ degrees
Area is $\frac{1}{2} \times 9 \times 5 \times sin (104.8) = 21.7$

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In figure, AB=9,BC=5, and ∠C=50∘. Calculate the area of triangle ABC, to the nearest tenth.

The correct option is A 21.7By sine rule, we get ABsin50=BCsinA ⇒sinA=59×sin50=0.426⇒A=25.2 degreesTherefore B=180−50−25.2=104.8 degreesArea is 12×9×5×sin(104.8)=21.7

In figure, $A B = 9, B C = 5$ , and $∠ C = 50^{\circ}$ . Calculate the area of triangle $A B C$ , to the nearest tenth.

The correct option is A $21.7$
By sine rule, we get
$\frac{A B}{sin 50} = \frac{B C}{sin A}$
$\Rightarrow sin A = \frac{5}{9} \times sin 50 = 0.426$
$\Rightarrow A = 25.2$ degrees
Therefore $B = 180 - 50 - 25.2 = 104.8$ degrees
Area is $\frac{1}{2} \times 9 \times 5 \times sin (104.8) = 21.7$