Question Video: Using Right-Angled Triangle Trigonometry to Evaluate the Tangent Function of the Sum of Two Angles in a Triangle | Nagwa Question Video: Using Right-Angled Triangle Trigonometry to Evaluate the Tangent Function of the Sum of Two Angles in a Triangle | Nagwa

# Question Video: Using Right-Angled Triangle Trigonometry to Evaluate the Tangent Function of the Sum of Two Angles in a Triangle Mathematics • Second Year of Secondary School

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The diagram shows triangle π΄π΅πΆ. Given that π΄π· is perpendicular to π΅πΆ, π΄π· = 15 cm, π΅π· = 10 cm, and πΆπ· = 7 cm, find the value of tan (π₯ + π¦).

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### Video Transcript

The diagram shows triangle π΄π΅πΆ. Given that π΄π· is perpendicular to π΅πΆ, π΄π· equals 15 centimeters, π΅π· equals 10 centimeters, and πΆπ· equals seven centimeters, find the value of the tan of π₯ plus π¦.

The first thing we want to do is label our diagram. π΄π· equals 15 centimeters, π΅π· equals 10 centimeters, and πΆπ· equals seven centimeters. Weβre interested in the tangent of this angle, the angle π₯ plus π¦. However, this angle is in a triangle thatβs not a right triangle, so weβll need a different strategy to solve for this angle. If we look, we see that angle π₯ and angle π¦ are both located inside right triangles. This means itβs possible to find the tangent value of π₯ and the tangent value of π¦. And by our angle sum identity, we know that the tan of π΄ plus π΅ will be equal to the tan of π΄ plus the tan of π΅ all over one minus the tan of π΄ times the tan of π΅.

The tangent of any angle in a right triangle will be equal to the opposite side length over the adjacent side length. For angle π₯, the opposite side length is 10 and the adjacent side length is 15. The tan of π₯ equals 10 over 15. To find the tan of π¦, we have the opposite side length of seven over the adjacent side length of 15. So the tan of π¦ equals seven fifteenths. So the tan of π₯ plus π¦ is equal to 10 over 15 plus seven over 15 all over one minus 10 over 15 times seven over 15. Our numerator becomes 17 over 15.

Before we do this multiplying in the denominator, we can do some simplifying. 10 over 15 simplifies to two-thirds, and two-thirds times seven fifteenths equals 14 over 45. In our denominator, since we have one minus 14 over 45, we can rewrite the one as 45 over 45. 45 minus 14 is 31. This means weβre dividing seventeen fifteenths by 31 over 45. And to divide by a fraction, we multiply by the reciprocal. We have seventeen fifteenths times 45 over 31. 15 goes into 45 three times. 17 times three equals 51, making the tan of π₯ plus π¦ 51 over 31.

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