.If two adjacent angles of a parallelogram are (3x 4 and (3x 10 then find the value of x))

\[\text{ We know that the adjacent angles of a parallelogram are supplementry } . \]

\[\text{ Hence }, \left( 3x + 10 \right)° \text{ and } \left( 3x - 4 \right)° \text{ are supplementry } . \]

\[\left( 3x + 10 \right)°+ \left( 3x - 4 \right)° = 180°\]

\[6x° + 6°= 180°\]

\[6x° = 174°\]

\[x = 29°\]

\[\text{ First angle } = \left( 3x + 10 \right)°= \left( 3 \times 29° + 10° \right) = 97°\]

\[\text{ Second angle }= \left( 3x - 4 \right)° = 83°\]

\[\text{ Thus, the angles of the parallelogram are } 97°, 83°, 97° \text{ and } 83°.\]

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Two adjacent angles of a parallelogram are 3 x 4∘ and 3 x+16∘. Find the value of x and hence find the measure of each of its angles.

Solution

In || gm ABCD, $\angle A\angle B$ are two adjacent angles

Let $\angle A=(3x-4{)}^{\circ }and\angle B=(3x+16{)}^{\circ }$

But $\angle A+\angle B={180}^{\circ }$

$\Rightarrow (3x-4{)}^{\circ }+(3x+16)={180}^{\circ }\Rightarrow 3x-4^{\circ }+3x+16={180}^{\circ }\Rightarrow 6x+{12}^{\circ }={180}^{\circ }\Rightarrow 6x={180}^{\circ }-{12}^{\circ }\Rightarrow 6x=168\Rightarrow x=\frac{168}{6}={28}^{\circ }\therefore x={28}^{\circ }Now\angle A=3x-4=3\times {28}^{\circ }-4^{\circ }={84}^{\circ }-4^{\circ }={80}^{\circ }\angle B=3x+16=3\times 28+16={84}^{\circ }+{16}^{\circ }={100}^{\circ }But\angle C=\angle A\text{opposite angles of}||gm\therefore \angle C={80}^{\circ }Similarly\angle D=\angle B={100}^{\circ }Hence\angle A={80}^{\circ },\angle B={100}^{\circ },\angle C={80}^{\circ }and\angle D={100}^{\circ }$

Mathematics

Secondary School Mathematics VIII

Standard VIII